Mixing
The difference between log and ln
$begingroup$
$$dfrac{1}{2}ln(x+7)-(2 ln x+3 ln y)$$
Our professor let's us solve this but i do not understand how $ln$ works. He says it has same properties with $log$ but i still don't get it. What's the difference of the two?
logarithms
edited Feb 12 '12 at 20:25
David
1,76252536
asked Dec 11 '11 at 22:17
ZhiancZhianc
213126
$endgroup$
|
show 5 more comments
$begingroup$
$$dfrac{1}{2}ln(x+7)-(2 ln x+3 ln y)$$
Our professor let's us solve this but i do not understand how $ln$ works. He says it has same properties with $log$ but i still don't get it. What's the difference of the two?
logarithms
edited Feb 12 '12 at 20:25
David
1,76252536
asked Dec 11 '11 at 22:17
ZhiancZhianc
213126
$endgroup$
8
$begingroup$
ln denotes the natural log, which has base $e$. Just "log" in this context denotes a log with base $10$. Google "mathematical constant e" for more info.
$endgroup$
– The Chaz 2.0
Dec 11 '11 at 22:20
5
$begingroup$
The formula is ambiguous. Do you mean $$frac{1}{2}ln(x+7)text{ or }frac{1}{2ln(x+7)}$$in the first summand?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:23
2
$begingroup$
@MaX: How do you know which one was meant?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:38
7
$begingroup$
The difference is that $ln$ is for children, and $log$ is for grownups.
$endgroup$
– Gerry Myerson
Dec 11 '11 at 22:46
7
$begingroup$
@GerryMyerson : You be nice or I'll whack you across the knuckles with my slide rule.
$endgroup$
– steven gregory
Nov 3 '15 at 18:39
|
show 5 more comments
$begingroup$
$$dfrac{1}{2}ln(x+7)-(2 ln x+3 ln y)$$
Our professor let's us solve this but i do not understand how $ln$ works. He says it has same properties with $log$ but i still don't get it. What's the difference of the two?
logarithms
edited Feb 12 '12 at 20:25
David
1,76252536
asked Dec 11 '11 at 22:17
ZhiancZhianc
213126
$endgroup$
$$dfrac{1}{2}ln(x+7)-(2 ln x+3 ln y)$$
Our professor let's us solve this but i do not understand how $ln$ works. He says it has same properties with $log$ but i still don't get it. What's the difference of the two?
logarithms
logarithms
edited Feb 12 '12 at 20:25
David
1,76252536
asked Dec 11 '11 at 22:17
ZhiancZhianc
213126
edited Feb 12 '12 at 20:25
David
1,76252536
asked Dec 11 '11 at 22:17
ZhiancZhianc
213126
edited Feb 12 '12 at 20:25
David
1,76252536
edited Feb 12 '12 at 20:25
David
1,76252536
edited Feb 12 '12 at 20:25
David
1,76252536
1,76252536
asked Dec 11 '11 at 22:17
ZhiancZhianc
213126
asked Dec 11 '11 at 22:17
ZhiancZhianc
213126
asked Dec 11 '11 at 22:17
ZhiancZhianc
213126
213126
8
$begingroup$
ln denotes the natural log, which has base $e$. Just "log" in this context denotes a log with base $10$. Google "mathematical constant e" for more info.
$endgroup$
– The Chaz 2.0
Dec 11 '11 at 22:20
5
$begingroup$
The formula is ambiguous. Do you mean $$frac{1}{2}ln(x+7)text{ or }frac{1}{2ln(x+7)}$$in the first summand?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:23
2
$begingroup$
@MaX: How do you know which one was meant?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:38
7
$begingroup$
The difference is that $ln$ is for children, and $log$ is for grownups.
$endgroup$
– Gerry Myerson
Dec 11 '11 at 22:46
7
$begingroup$
@GerryMyerson : You be nice or I'll whack you across the knuckles with my slide rule.
$endgroup$
– steven gregory
Nov 3 '15 at 18:39
|
show 5 more comments
8
$begingroup$
ln denotes the natural log, which has base $e$. Just "log" in this context denotes a log with base $10$. Google "mathematical constant e" for more info.
$endgroup$
– The Chaz 2.0
Dec 11 '11 at 22:20
5
$begingroup$
The formula is ambiguous. Do you mean $$frac{1}{2}ln(x+7)text{ or }frac{1}{2ln(x+7)}$$in the first summand?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:23
2
$begingroup$
@MaX: How do you know which one was meant?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:38
7
$begingroup$
The difference is that $ln$ is for children, and $log$ is for grownups.
$endgroup$
– Gerry Myerson
Dec 11 '11 at 22:46
7
$begingroup$
@GerryMyerson : You be nice or I'll whack you across the knuckles with my slide rule.
$endgroup$
– steven gregory
Nov 3 '15 at 18:39
8
8
$begingroup$
ln denotes the natural log, which has base $e$. Just "log" in this context denotes a log with base $10$. Google "mathematical constant e" for more info.
$endgroup$
– The Chaz 2.0
Dec 11 '11 at 22:20
$begingroup$
ln denotes the natural log, which has base $e$. Just "log" in this context denotes a log with base $10$. Google "mathematical constant e" for more info.
$endgroup$
– The Chaz 2.0
Dec 11 '11 at 22:20
5
5
$begingroup$
The formula is ambiguous. Do you mean $$frac{1}{2}ln(x+7)text{ or }frac{1}{2ln(x+7)}$$in the first summand?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:23
$begingroup$
The formula is ambiguous. Do you mean $$frac{1}{2}ln(x+7)text{ or }frac{1}{2ln(x+7)}$$in the first summand?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:23
2
2
$begingroup$
@MaX: How do you know which one was meant?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:38
$begingroup$
@MaX: How do you know which one was meant?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:38
7
7
$begingroup$
The difference is that $ln$ is for children, and $log$ is for grownups.
$endgroup$
– Gerry Myerson
Dec 11 '11 at 22:46
$begingroup$
The difference is that $ln$ is for children, and $log$ is for grownups.
$endgroup$
– Gerry Myerson
Dec 11 '11 at 22:46
7
7
$begingroup$
@GerryMyerson : You be nice or I'll whack you across the knuckles with my slide rule.
$endgroup$
– steven gregory
Nov 3 '15 at 18:39
$begingroup$
@GerryMyerson : You be nice or I'll whack you across the knuckles with my slide rule.
$endgroup$
– steven gregory
Nov 3 '15 at 18:39
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The common logarithm, is the logarithm base 10. It is the inverse of the exponential function $10^x$. In Calculus and Precalculus classes, it is usually denoted $log$.
The natural logarithm, is the logarithm base $e$. It is the inverse of the exponential function $e^x$. In Calculus and Precalculus classes, it is often denoted $ln$.
In general, if $agt 0$, $aneq 1$, then the inverse of the function $a^x$ is the "logarithm base $a$", $log_a(x)$.
The "guiding formula" is
$$log_a(b) = rtext{ if and only if }a^r = b.$$
From these, the properties of the logarithmic functions follow:
$log_a(xy) = log_a(x)+log_a(y)$: logarithm of a product is the sum of the logarithms.
Why? Say $log_a(x) = r$ and $log_a(y)=s$. That means that $a^r = x$ and $a^s=y$. Then $xy = a^ra^s = a^{r+s}$, so $log_a(xy) = r+s = log_a(x) + log_a(y)$.
$log_aleft(frac{x}{y}right) = log_a(x) - log_a(y)$.
Why? Again, say $log_a(x) = r$ and $log_a(y) = s$. Then $a^r = x$, $a^s = y$, so
$frac{x}{y} = frac{a^r}{a^s} = a^{r-s}$, which means $log_afrac{x}{y}=r-s = log_a(x)-log_a(y)$.
$log_a(x^t) = tlog_a(x)$$.
Why? If $log_a(x)=r$, so that $a^r = x$, then $x^t = (a^r)^t = a^{rt}$, so $log_a(x^t) = rt = tlog_a(x)$.
$log_a(a^r) = r$ and $a^{log_a(x)} = x$. Because $log_a(x)$ and $a^x$ are inverses of each other.
In particular, $ln$, which is $log_{e}$; and using $log$ for $log_{10}$, we have these properties:
$$begin{align*}
log(xy) &= log(x)+log(y) &qquad ln(xy) &=ln(x) + ln(y)\
logleft(frac{x}{y}right) &= log(x) - log(y) &lnleft(frac{x}{y}right) &= ln(x) - ln(y)\
log(x^a) &= alog(x) & ln(x^a) &= aln(x)\
log(10^x) &= x & ln(e^x) &= x\
10^{log(x)} &= x & e^{ln(x)} &= x
end{align*}$$
It also gives you a way to go back and forth between any logarithm and any other logarithm: if $a$ and $b$ are two bases, both positive, both different from one, what is the relation between $log_a(x)$ and $log_b(x)$?
If $log_b(x)=r$, then $b^r = x$. So
$$log_a(x)= log_a(b^r) = rlog_a(b) = log_b(x)log_a(b).$$
So we get that
$$log_b(x) = frac{log_a(x)}{log_a(b)}.$$
As Henning points out below, while $ln$ is not ambiguous (it always denotes logarithm base $e$), $log$ is ambiguous and its exact meaning depends on context. In more advanced mathematics courses, it is usual to use it to mean the natural logarithm; in computer science it is very often used to denote logarithm base $2$. For some applications it does not matter (for example, when analyzing complexity, since two different logarithms are just scalar multiples of each other).
answered Dec 11 '11 at 22:31
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
22
$begingroup$
Beware that $log$ does not unambiguously mean the base-10 logarithm, but rather "the logarithm that we usually use". In many areas of higher mathematics, $log$ means the natural logarithm and the $ln$ notation is seldom seen. And computer scientists routinely use $log$ to mean $log_2$.
$endgroup$
– Henning Makholm
Dec 11 '11 at 22:49
1
$begingroup$
Good point; I guess I've been teaching too many Calculii lately...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:59
2
$begingroup$
Don't even assume that log = base2 for software; TensorFlow, for example, has a log() function that "Computes natural logarithm of x element-wise." The use of "log" appears to be intentionally annoying across fields.
$endgroup$
– James Moore
Nov 4 '16 at 15:38
$begingroup$
Can I say that, logarithmic function does not have its own definition except that it is originated as inverse of exponential(e^x) function?
$endgroup$
– overexchange
Nov 24 '16 at 13:34
add a comment |
$begingroup$
The use of the "ln" abbreviation for natural logarithm is a bad thing because it makes people think that "log" is one thing and "ln" is another thing, and ask what's the difference between the two.
The base-$10$ logarithmic function is a logarithmic function.
The base-$2$ logarithmic function is a logarithmic function.
The base-$e$ logarithmic function is a logarithmic function.
The difference is which number is the base.
Mathematicians writing "$log x$" usually mean $log_e x$, also called $ln x$.
Calculators use $log x$ to mean $log_{10} x$. This is also used in some of the sciences when doing numerical things.
The reason for the importance of base-$10$ logarithms was made obsolete by calculators. In the early '70s, calculators became widespread. Before then, many books had tables of base-$10$ logarithms in an appendix. Suppose you wanted the logarithm of $123$ The table gave you logarithms of numbers between $1$ and $10$, so you found $log_{10}1.23= 0.089905ldots$ and concluded that $log_{10} 123 = 2.089905ldots;{}$. You added $2$ to move the decimal point over 2 places. That's why base 10 was used: to make that possible. If you wanted the square root of $7$, you found the logarithm of $7$, divided by $2$, then found the antilogarithm in the same table. If you wanted to divide $319450231$ by $2673019201$, you found logarithms of both in the table, subtracted, and then found the antilogarithm. And so on.
The important theoretical question to ask about "$ln$" is why $e=2.71828182846ldots$ is the "natural" base to use. (Has someone posted that question here?) (When I raise that question and try to answer it in a calculus class, some students ask "Do we HAVE to know this?? Will it be on the test?". Next time someone does that, I'm going to say "Who cares?".)
answered Dec 11 '11 at 23:10
Michael HardyMichael Hardy
1
$endgroup$
5
$begingroup$
I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 23:25
3
$begingroup$
"The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here.
$endgroup$
– Austin Mohr
Dec 12 '11 at 0:21
add a comment |
$begingroup$
Logs with different bases cross the line y = 0 at x=1 with different slopes (of the tangent to the curve). Natural base e makes this slope equal to 1.
The number is called e after Leonhard Euler, a mathematician that first gave this number a meaning and found its value. Euler worked on a formula for compounding interest. If r is the annualized interest rate and n is a number of compounding intervals per year, the formula for the amount of investment of $1 after n intervals is: $$(1+frac r n )^n$$ Euler showed that the limit of this value for infinitely large n is $e^r$ where e is $$lim_{ntoinfty} (1+frac 1 n)^n$$ when $n rightarrow infty$. It is approximately 2.718.
edited Nov 21 '16 at 0:04
xtiansimon
1546
answered Aug 8 '16 at 16:30
Valery FradkovValery Fradkov
111
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
The common logarithm, is the logarithm base 10. It is the inverse of the exponential function $10^x$. In Calculus and Precalculus classes, it is usually denoted $log$.
The natural logarithm, is the logarithm base $e$. It is the inverse of the exponential function $e^x$. In Calculus and Precalculus classes, it is often denoted $ln$.
In general, if $agt 0$, $aneq 1$, then the inverse of the function $a^x$ is the "logarithm base $a$", $log_a(x)$.
The "guiding formula" is
$$log_a(b) = rtext{ if and only if }a^r = b.$$
From these, the properties of the logarithmic functions follow:
$log_a(xy) = log_a(x)+log_a(y)$: logarithm of a product is the sum of the logarithms.
Why? Say $log_a(x) = r$ and $log_a(y)=s$. That means that $a^r = x$ and $a^s=y$. Then $xy = a^ra^s = a^{r+s}$, so $log_a(xy) = r+s = log_a(x) + log_a(y)$.
$log_aleft(frac{x}{y}right) = log_a(x) - log_a(y)$.
Why? Again, say $log_a(x) = r$ and $log_a(y) = s$. Then $a^r = x$, $a^s = y$, so
$frac{x}{y} = frac{a^r}{a^s} = a^{r-s}$, which means $log_afrac{x}{y}=r-s = log_a(x)-log_a(y)$.
$log_a(x^t) = tlog_a(x)$$.
Why? If $log_a(x)=r$, so that $a^r = x$, then $x^t = (a^r)^t = a^{rt}$, so $log_a(x^t) = rt = tlog_a(x)$.
$log_a(a^r) = r$ and $a^{log_a(x)} = x$. Because $log_a(x)$ and $a^x$ are inverses of each other.
In particular, $ln$, which is $log_{e}$; and using $log$ for $log_{10}$, we have these properties:
$$begin{align*}
log(xy) &= log(x)+log(y) &qquad ln(xy) &=ln(x) + ln(y)\
logleft(frac{x}{y}right) &= log(x) - log(y) &lnleft(frac{x}{y}right) &= ln(x) - ln(y)\
log(x^a) &= alog(x) & ln(x^a) &= aln(x)\
log(10^x) &= x & ln(e^x) &= x\
10^{log(x)} &= x & e^{ln(x)} &= x
end{align*}$$
It also gives you a way to go back and forth between any logarithm and any other logarithm: if $a$ and $b$ are two bases, both positive, both different from one, what is the relation between $log_a(x)$ and $log_b(x)$?
If $log_b(x)=r$, then $b^r = x$. So
$$log_a(x)= log_a(b^r) = rlog_a(b) = log_b(x)log_a(b).$$
So we get that
$$log_b(x) = frac{log_a(x)}{log_a(b)}.$$
As Henning points out below, while $ln$ is not ambiguous (it always denotes logarithm base $e$), $log$ is ambiguous and its exact meaning depends on context. In more advanced mathematics courses, it is usual to use it to mean the natural logarithm; in computer science it is very often used to denote logarithm base $2$. For some applications it does not matter (for example, when analyzing complexity, since two different logarithms are just scalar multiples of each other).
answered Dec 11 '11 at 22:31
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
22
$begingroup$
Beware that $log$ does not unambiguously mean the base-10 logarithm, but rather "the logarithm that we usually use". In many areas of higher mathematics, $log$ means the natural logarithm and the $ln$ notation is seldom seen. And computer scientists routinely use $log$ to mean $log_2$.
$endgroup$
– Henning Makholm
Dec 11 '11 at 22:49
1
$begingroup$
Good point; I guess I've been teaching too many Calculii lately...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:59
2
$begingroup$
Don't even assume that log = base2 for software; TensorFlow, for example, has a log() function that "Computes natural logarithm of x element-wise." The use of "log" appears to be intentionally annoying across fields.
$endgroup$
– James Moore
Nov 4 '16 at 15:38
$begingroup$
Can I say that, logarithmic function does not have its own definition except that it is originated as inverse of exponential(e^x) function?
$endgroup$
– overexchange
Nov 24 '16 at 13:34
add a comment |
$begingroup$
The common logarithm, is the logarithm base 10. It is the inverse of the exponential function $10^x$. In Calculus and Precalculus classes, it is usually denoted $log$.
The natural logarithm, is the logarithm base $e$. It is the inverse of the exponential function $e^x$. In Calculus and Precalculus classes, it is often denoted $ln$.
In general, if $agt 0$, $aneq 1$, then the inverse of the function $a^x$ is the "logarithm base $a$", $log_a(x)$.
The "guiding formula" is
$$log_a(b) = rtext{ if and only if }a^r = b.$$
From these, the properties of the logarithmic functions follow:
$log_a(xy) = log_a(x)+log_a(y)$: logarithm of a product is the sum of the logarithms.
Why? Say $log_a(x) = r$ and $log_a(y)=s$. That means that $a^r = x$ and $a^s=y$. Then $xy = a^ra^s = a^{r+s}$, so $log_a(xy) = r+s = log_a(x) + log_a(y)$.
$log_aleft(frac{x}{y}right) = log_a(x) - log_a(y)$.
Why? Again, say $log_a(x) = r$ and $log_a(y) = s$. Then $a^r = x$, $a^s = y$, so
$frac{x}{y} = frac{a^r}{a^s} = a^{r-s}$, which means $log_afrac{x}{y}=r-s = log_a(x)-log_a(y)$.
$log_a(x^t) = tlog_a(x)$$.
Why? If $log_a(x)=r$, so that $a^r = x$, then $x^t = (a^r)^t = a^{rt}$, so $log_a(x^t) = rt = tlog_a(x)$.
$log_a(a^r) = r$ and $a^{log_a(x)} = x$. Because $log_a(x)$ and $a^x$ are inverses of each other.
In particular, $ln$, which is $log_{e}$; and using $log$ for $log_{10}$, we have these properties:
$$begin{align*}
log(xy) &= log(x)+log(y) &qquad ln(xy) &=ln(x) + ln(y)\
logleft(frac{x}{y}right) &= log(x) - log(y) &lnleft(frac{x}{y}right) &= ln(x) - ln(y)\
log(x^a) &= alog(x) & ln(x^a) &= aln(x)\
log(10^x) &= x & ln(e^x) &= x\
10^{log(x)} &= x & e^{ln(x)} &= x
end{align*}$$
It also gives you a way to go back and forth between any logarithm and any other logarithm: if $a$ and $b$ are two bases, both positive, both different from one, what is the relation between $log_a(x)$ and $log_b(x)$?
If $log_b(x)=r$, then $b^r = x$. So
$$log_a(x)= log_a(b^r) = rlog_a(b) = log_b(x)log_a(b).$$
So we get that
$$log_b(x) = frac{log_a(x)}{log_a(b)}.$$
As Henning points out below, while $ln$ is not ambiguous (it always denotes logarithm base $e$), $log$ is ambiguous and its exact meaning depends on context. In more advanced mathematics courses, it is usual to use it to mean the natural logarithm; in computer science it is very often used to denote logarithm base $2$. For some applications it does not matter (for example, when analyzing complexity, since two different logarithms are just scalar multiples of each other).
answered Dec 11 '11 at 22:31
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
22
$begingroup$
Beware that $log$ does not unambiguously mean the base-10 logarithm, but rather "the logarithm that we usually use". In many areas of higher mathematics, $log$ means the natural logarithm and the $ln$ notation is seldom seen. And computer scientists routinely use $log$ to mean $log_2$.
$endgroup$
– Henning Makholm
Dec 11 '11 at 22:49
1
$begingroup$
Good point; I guess I've been teaching too many Calculii lately...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:59
2
$begingroup$
Don't even assume that log = base2 for software; TensorFlow, for example, has a log() function that "Computes natural logarithm of x element-wise." The use of "log" appears to be intentionally annoying across fields.
$endgroup$
– James Moore
Nov 4 '16 at 15:38
$begingroup$
Can I say that, logarithmic function does not have its own definition except that it is originated as inverse of exponential(e^x) function?
$endgroup$
– overexchange
Nov 24 '16 at 13:34
add a comment |
$begingroup$
The common logarithm, is the logarithm base 10. It is the inverse of the exponential function $10^x$. In Calculus and Precalculus classes, it is usually denoted $log$.
The natural logarithm, is the logarithm base $e$. It is the inverse of the exponential function $e^x$. In Calculus and Precalculus classes, it is often denoted $ln$.
In general, if $agt 0$, $aneq 1$, then the inverse of the function $a^x$ is the "logarithm base $a$", $log_a(x)$.
The "guiding formula" is
$$log_a(b) = rtext{ if and only if }a^r = b.$$
From these, the properties of the logarithmic functions follow:
$log_a(xy) = log_a(x)+log_a(y)$: logarithm of a product is the sum of the logarithms.
Why? Say $log_a(x) = r$ and $log_a(y)=s$. That means that $a^r = x$ and $a^s=y$. Then $xy = a^ra^s = a^{r+s}$, so $log_a(xy) = r+s = log_a(x) + log_a(y)$.
$log_aleft(frac{x}{y}right) = log_a(x) - log_a(y)$.
Why? Again, say $log_a(x) = r$ and $log_a(y) = s$. Then $a^r = x$, $a^s = y$, so
$frac{x}{y} = frac{a^r}{a^s} = a^{r-s}$, which means $log_afrac{x}{y}=r-s = log_a(x)-log_a(y)$.
$log_a(x^t) = tlog_a(x)$$.
Why? If $log_a(x)=r$, so that $a^r = x$, then $x^t = (a^r)^t = a^{rt}$, so $log_a(x^t) = rt = tlog_a(x)$.
$log_a(a^r) = r$ and $a^{log_a(x)} = x$. Because $log_a(x)$ and $a^x$ are inverses of each other.
In particular, $ln$, which is $log_{e}$; and using $log$ for $log_{10}$, we have these properties:
$$begin{align*}
log(xy) &= log(x)+log(y) &qquad ln(xy) &=ln(x) + ln(y)\
logleft(frac{x}{y}right) &= log(x) - log(y) &lnleft(frac{x}{y}right) &= ln(x) - ln(y)\
log(x^a) &= alog(x) & ln(x^a) &= aln(x)\
log(10^x) &= x & ln(e^x) &= x\
10^{log(x)} &= x & e^{ln(x)} &= x
end{align*}$$
It also gives you a way to go back and forth between any logarithm and any other logarithm: if $a$ and $b$ are two bases, both positive, both different from one, what is the relation between $log_a(x)$ and $log_b(x)$?
If $log_b(x)=r$, then $b^r = x$. So
$$log_a(x)= log_a(b^r) = rlog_a(b) = log_b(x)log_a(b).$$
So we get that
$$log_b(x) = frac{log_a(x)}{log_a(b)}.$$
As Henning points out below, while $ln$ is not ambiguous (it always denotes logarithm base $e$), $log$ is ambiguous and its exact meaning depends on context. In more advanced mathematics courses, it is usual to use it to mean the natural logarithm; in computer science it is very often used to denote logarithm base $2$. For some applications it does not matter (for example, when analyzing complexity, since two different logarithms are just scalar multiples of each other).
answered Dec 11 '11 at 22:31
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
The common logarithm, is the logarithm base 10. It is the inverse of the exponential function $10^x$. In Calculus and Precalculus classes, it is usually denoted $log$.
The natural logarithm, is the logarithm base $e$. It is the inverse of the exponential function $e^x$. In Calculus and Precalculus classes, it is often denoted $ln$.
In general, if $agt 0$, $aneq 1$, then the inverse of the function $a^x$ is the "logarithm base $a$", $log_a(x)$.
The "guiding formula" is
$$log_a(b) = rtext{ if and only if }a^r = b.$$
From these, the properties of the logarithmic functions follow:
$log_a(xy) = log_a(x)+log_a(y)$: logarithm of a product is the sum of the logarithms.
Why? Say $log_a(x) = r$ and $log_a(y)=s$. That means that $a^r = x$ and $a^s=y$. Then $xy = a^ra^s = a^{r+s}$, so $log_a(xy) = r+s = log_a(x) + log_a(y)$.
$log_aleft(frac{x}{y}right) = log_a(x) - log_a(y)$.
Why? Again, say $log_a(x) = r$ and $log_a(y) = s$. Then $a^r = x$, $a^s = y$, so
$frac{x}{y} = frac{a^r}{a^s} = a^{r-s}$, which means $log_afrac{x}{y}=r-s = log_a(x)-log_a(y)$.
$log_a(x^t) = tlog_a(x)$$.
Why? If $log_a(x)=r$, so that $a^r = x$, then $x^t = (a^r)^t = a^{rt}$, so $log_a(x^t) = rt = tlog_a(x)$.
$log_a(a^r) = r$ and $a^{log_a(x)} = x$. Because $log_a(x)$ and $a^x$ are inverses of each other.
In particular, $ln$, which is $log_{e}$; and using $log$ for $log_{10}$, we have these properties:
$$begin{align*}
log(xy) &= log(x)+log(y) &qquad ln(xy) &=ln(x) + ln(y)\
logleft(frac{x}{y}right) &= log(x) - log(y) &lnleft(frac{x}{y}right) &= ln(x) - ln(y)\
log(x^a) &= alog(x) & ln(x^a) &= aln(x)\
log(10^x) &= x & ln(e^x) &= x\
10^{log(x)} &= x & e^{ln(x)} &= x
end{align*}$$
It also gives you a way to go back and forth between any logarithm and any other logarithm: if $a$ and $b$ are two bases, both positive, both different from one, what is the relation between $log_a(x)$ and $log_b(x)$?
If $log_b(x)=r$, then $b^r = x$. So
$$log_a(x)= log_a(b^r) = rlog_a(b) = log_b(x)log_a(b).$$
So we get that
$$log_b(x) = frac{log_a(x)}{log_a(b)}.$$
As Henning points out below, while $ln$ is not ambiguous (it always denotes logarithm base $e$), $log$ is ambiguous and its exact meaning depends on context. In more advanced mathematics courses, it is usual to use it to mean the natural logarithm; in computer science it is very often used to denote logarithm base $2$. For some applications it does not matter (for example, when analyzing complexity, since two different logarithms are just scalar multiples of each other).
answered Dec 11 '11 at 22:31
Arturo MagidinArturo Magidin
263k34587911
edited Dec 11 '11 at 23:03
answered Dec 11 '11 at 22:31
Arturo MagidinArturo Magidin
263k34587911
answered Dec 11 '11 at 22:31
Arturo MagidinArturo Magidin
263k34587911
answered Dec 11 '11 at 22:31
Arturo MagidinArturo Magidin
263k34587911
263k34587911
22
$begingroup$
Beware that $log$ does not unambiguously mean the base-10 logarithm, but rather "the logarithm that we usually use". In many areas of higher mathematics, $log$ means the natural logarithm and the $ln$ notation is seldom seen. And computer scientists routinely use $log$ to mean $log_2$.
$endgroup$
– Henning Makholm
Dec 11 '11 at 22:49
1
$begingroup$
Good point; I guess I've been teaching too many Calculii lately...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:59
2
$begingroup$
Don't even assume that log = base2 for software; TensorFlow, for example, has a log() function that "Computes natural logarithm of x element-wise." The use of "log" appears to be intentionally annoying across fields.
$endgroup$
– James Moore
Nov 4 '16 at 15:38
$begingroup$
Can I say that, logarithmic function does not have its own definition except that it is originated as inverse of exponential(e^x) function?
$endgroup$
– overexchange
Nov 24 '16 at 13:34
add a comment |
22
$begingroup$
Beware that $log$ does not unambiguously mean the base-10 logarithm, but rather "the logarithm that we usually use". In many areas of higher mathematics, $log$ means the natural logarithm and the $ln$ notation is seldom seen. And computer scientists routinely use $log$ to mean $log_2$.
$endgroup$
– Henning Makholm
Dec 11 '11 at 22:49
1
$begingroup$
Good point; I guess I've been teaching too many Calculii lately...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:59
2
$begingroup$
Don't even assume that log = base2 for software; TensorFlow, for example, has a log() function that "Computes natural logarithm of x element-wise." The use of "log" appears to be intentionally annoying across fields.
$endgroup$
– James Moore
Nov 4 '16 at 15:38
$begingroup$
Can I say that, logarithmic function does not have its own definition except that it is originated as inverse of exponential(e^x) function?
$endgroup$
– overexchange
Nov 24 '16 at 13:34
22
22
$begingroup$
Beware that $log$ does not unambiguously mean the base-10 logarithm, but rather "the logarithm that we usually use". In many areas of higher mathematics, $log$ means the natural logarithm and the $ln$ notation is seldom seen. And computer scientists routinely use $log$ to mean $log_2$.
$endgroup$
– Henning Makholm
Dec 11 '11 at 22:49
$begingroup$
Beware that $log$ does not unambiguously mean the base-10 logarithm, but rather "the logarithm that we usually use". In many areas of higher mathematics, $log$ means the natural logarithm and the $ln$ notation is seldom seen. And computer scientists routinely use $log$ to mean $log_2$.
$endgroup$
– Henning Makholm
Dec 11 '11 at 22:49
1
1
$begingroup$
Good point; I guess I've been teaching too many Calculii lately...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:59
$begingroup$
Good point; I guess I've been teaching too many Calculii lately...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:59
2
2
$begingroup$
Don't even assume that log = base2 for software; TensorFlow, for example, has a log() function that "Computes natural logarithm of x element-wise." The use of "log" appears to be intentionally annoying across fields.
$endgroup$
– James Moore
Nov 4 '16 at 15:38
$begingroup$
Don't even assume that log = base2 for software; TensorFlow, for example, has a log() function that "Computes natural logarithm of x element-wise." The use of "log" appears to be intentionally annoying across fields.
$endgroup$
– James Moore
Nov 4 '16 at 15:38
$begingroup$
Can I say that, logarithmic function does not have its own definition except that it is originated as inverse of exponential(e^x) function?
$endgroup$
– overexchange
Nov 24 '16 at 13:34
$begingroup$
Can I say that, logarithmic function does not have its own definition except that it is originated as inverse of exponential(e^x) function?
$endgroup$
– overexchange
Nov 24 '16 at 13:34
add a comment |
$begingroup$
The use of the "ln" abbreviation for natural logarithm is a bad thing because it makes people think that "log" is one thing and "ln" is another thing, and ask what's the difference between the two.
The base-$10$ logarithmic function is a logarithmic function.
The base-$2$ logarithmic function is a logarithmic function.
The base-$e$ logarithmic function is a logarithmic function.
The difference is which number is the base.
Mathematicians writing "$log x$" usually mean $log_e x$, also called $ln x$.
Calculators use $log x$ to mean $log_{10} x$. This is also used in some of the sciences when doing numerical things.
The reason for the importance of base-$10$ logarithms was made obsolete by calculators. In the early '70s, calculators became widespread. Before then, many books had tables of base-$10$ logarithms in an appendix. Suppose you wanted the logarithm of $123$ The table gave you logarithms of numbers between $1$ and $10$, so you found $log_{10}1.23= 0.089905ldots$ and concluded that $log_{10} 123 = 2.089905ldots;{}$. You added $2$ to move the decimal point over 2 places. That's why base 10 was used: to make that possible. If you wanted the square root of $7$, you found the logarithm of $7$, divided by $2$, then found the antilogarithm in the same table. If you wanted to divide $319450231$ by $2673019201$, you found logarithms of both in the table, subtracted, and then found the antilogarithm. And so on.
The important theoretical question to ask about "$ln$" is why $e=2.71828182846ldots$ is the "natural" base to use. (Has someone posted that question here?) (When I raise that question and try to answer it in a calculus class, some students ask "Do we HAVE to know this?? Will it be on the test?". Next time someone does that, I'm going to say "Who cares?".)
answered Dec 11 '11 at 23:10
Michael HardyMichael Hardy
1
$endgroup$
5
$begingroup$
I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 23:25
3
$begingroup$
"The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here.
$endgroup$
– Austin Mohr
Dec 12 '11 at 0:21
add a comment |
$begingroup$
The use of the "ln" abbreviation for natural logarithm is a bad thing because it makes people think that "log" is one thing and "ln" is another thing, and ask what's the difference between the two.
The base-$10$ logarithmic function is a logarithmic function.
The base-$2$ logarithmic function is a logarithmic function.
The base-$e$ logarithmic function is a logarithmic function.
The difference is which number is the base.
Mathematicians writing "$log x$" usually mean $log_e x$, also called $ln x$.
Calculators use $log x$ to mean $log_{10} x$. This is also used in some of the sciences when doing numerical things.
The reason for the importance of base-$10$ logarithms was made obsolete by calculators. In the early '70s, calculators became widespread. Before then, many books had tables of base-$10$ logarithms in an appendix. Suppose you wanted the logarithm of $123$ The table gave you logarithms of numbers between $1$ and $10$, so you found $log_{10}1.23= 0.089905ldots$ and concluded that $log_{10} 123 = 2.089905ldots;{}$. You added $2$ to move the decimal point over 2 places. That's why base 10 was used: to make that possible. If you wanted the square root of $7$, you found the logarithm of $7$, divided by $2$, then found the antilogarithm in the same table. If you wanted to divide $319450231$ by $2673019201$, you found logarithms of both in the table, subtracted, and then found the antilogarithm. And so on.
The important theoretical question to ask about "$ln$" is why $e=2.71828182846ldots$ is the "natural" base to use. (Has someone posted that question here?) (When I raise that question and try to answer it in a calculus class, some students ask "Do we HAVE to know this?? Will it be on the test?". Next time someone does that, I'm going to say "Who cares?".)
answered Dec 11 '11 at 23:10
Michael HardyMichael Hardy
1
$endgroup$
5
$begingroup$
I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 23:25
3
$begingroup$
"The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here.
$endgroup$
– Austin Mohr
Dec 12 '11 at 0:21
add a comment |
$begingroup$
The use of the "ln" abbreviation for natural logarithm is a bad thing because it makes people think that "log" is one thing and "ln" is another thing, and ask what's the difference between the two.
The base-$10$ logarithmic function is a logarithmic function.
The base-$2$ logarithmic function is a logarithmic function.
The base-$e$ logarithmic function is a logarithmic function.
The difference is which number is the base.
Mathematicians writing "$log x$" usually mean $log_e x$, also called $ln x$.
Calculators use $log x$ to mean $log_{10} x$. This is also used in some of the sciences when doing numerical things.
The reason for the importance of base-$10$ logarithms was made obsolete by calculators. In the early '70s, calculators became widespread. Before then, many books had tables of base-$10$ logarithms in an appendix. Suppose you wanted the logarithm of $123$ The table gave you logarithms of numbers between $1$ and $10$, so you found $log_{10}1.23= 0.089905ldots$ and concluded that $log_{10} 123 = 2.089905ldots;{}$. You added $2$ to move the decimal point over 2 places. That's why base 10 was used: to make that possible. If you wanted the square root of $7$, you found the logarithm of $7$, divided by $2$, then found the antilogarithm in the same table. If you wanted to divide $319450231$ by $2673019201$, you found logarithms of both in the table, subtracted, and then found the antilogarithm. And so on.
The important theoretical question to ask about "$ln$" is why $e=2.71828182846ldots$ is the "natural" base to use. (Has someone posted that question here?) (When I raise that question and try to answer it in a calculus class, some students ask "Do we HAVE to know this?? Will it be on the test?". Next time someone does that, I'm going to say "Who cares?".)
answered Dec 11 '11 at 23:10
Michael HardyMichael Hardy
1
$endgroup$
The use of the "ln" abbreviation for natural logarithm is a bad thing because it makes people think that "log" is one thing and "ln" is another thing, and ask what's the difference between the two.
The base-$10$ logarithmic function is a logarithmic function.
The base-$2$ logarithmic function is a logarithmic function.
The base-$e$ logarithmic function is a logarithmic function.
The difference is which number is the base.
Mathematicians writing "$log x$" usually mean $log_e x$, also called $ln x$.
Calculators use $log x$ to mean $log_{10} x$. This is also used in some of the sciences when doing numerical things.
The reason for the importance of base-$10$ logarithms was made obsolete by calculators. In the early '70s, calculators became widespread. Before then, many books had tables of base-$10$ logarithms in an appendix. Suppose you wanted the logarithm of $123$ The table gave you logarithms of numbers between $1$ and $10$, so you found $log_{10}1.23= 0.089905ldots$ and concluded that $log_{10} 123 = 2.089905ldots;{}$. You added $2$ to move the decimal point over 2 places. That's why base 10 was used: to make that possible. If you wanted the square root of $7$, you found the logarithm of $7$, divided by $2$, then found the antilogarithm in the same table. If you wanted to divide $319450231$ by $2673019201$, you found logarithms of both in the table, subtracted, and then found the antilogarithm. And so on.
The important theoretical question to ask about "$ln$" is why $e=2.71828182846ldots$ is the "natural" base to use. (Has someone posted that question here?) (When I raise that question and try to answer it in a calculus class, some students ask "Do we HAVE to know this?? Will it be on the test?". Next time someone does that, I'm going to say "Who cares?".)
answered Dec 11 '11 at 23:10
Michael HardyMichael Hardy
1
answered Dec 11 '11 at 23:10
Michael HardyMichael Hardy
1
answered Dec 11 '11 at 23:10
Michael HardyMichael Hardy
1
answered Dec 11 '11 at 23:10
Michael HardyMichael Hardy
1
1
5
$begingroup$
I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 23:25
3
$begingroup$
"The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here.
$endgroup$
– Austin Mohr
Dec 12 '11 at 0:21
add a comment |
5
$begingroup$
I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 23:25
3
$begingroup$
"The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here.
$endgroup$
– Austin Mohr
Dec 12 '11 at 0:21
5
5
$begingroup$
I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 23:25
$begingroup$
I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"...
$endgroup$
– Arturo Magidin
Dec 11 '11 at 23:25
3
3
$begingroup$
"The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here.
$endgroup$
– Austin Mohr
Dec 12 '11 at 0:21
$begingroup$
"The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here.
$endgroup$
– Austin Mohr
Dec 12 '11 at 0:21
add a comment |
$begingroup$
Logs with different bases cross the line y = 0 at x=1 with different slopes (of the tangent to the curve). Natural base e makes this slope equal to 1.
The number is called e after Leonhard Euler, a mathematician that first gave this number a meaning and found its value. Euler worked on a formula for compounding interest. If r is the annualized interest rate and n is a number of compounding intervals per year, the formula for the amount of investment of $1 after n intervals is: $$(1+frac r n )^n$$ Euler showed that the limit of this value for infinitely large n is $e^r$ where e is $$lim_{ntoinfty} (1+frac 1 n)^n$$ when $n rightarrow infty$. It is approximately 2.718.
edited Nov 21 '16 at 0:04
xtiansimon
1546
answered Aug 8 '16 at 16:30
Valery FradkovValery Fradkov
111
$endgroup$
add a comment |
$begingroup$
Logs with different bases cross the line y = 0 at x=1 with different slopes (of the tangent to the curve). Natural base e makes this slope equal to 1.
The number is called e after Leonhard Euler, a mathematician that first gave this number a meaning and found its value. Euler worked on a formula for compounding interest. If r is the annualized interest rate and n is a number of compounding intervals per year, the formula for the amount of investment of $1 after n intervals is: $$(1+frac r n )^n$$ Euler showed that the limit of this value for infinitely large n is $e^r$ where e is $$lim_{ntoinfty} (1+frac 1 n)^n$$ when $n rightarrow infty$. It is approximately 2.718.
edited Nov 21 '16 at 0:04
xtiansimon
1546
answered Aug 8 '16 at 16:30
Valery FradkovValery Fradkov
111
$endgroup$
add a comment |
$begingroup$
Logs with different bases cross the line y = 0 at x=1 with different slopes (of the tangent to the curve). Natural base e makes this slope equal to 1.
The number is called e after Leonhard Euler, a mathematician that first gave this number a meaning and found its value. Euler worked on a formula for compounding interest. If r is the annualized interest rate and n is a number of compounding intervals per year, the formula for the amount of investment of $1 after n intervals is: $$(1+frac r n )^n$$ Euler showed that the limit of this value for infinitely large n is $e^r$ where e is $$lim_{ntoinfty} (1+frac 1 n)^n$$ when $n rightarrow infty$. It is approximately 2.718.
edited Nov 21 '16 at 0:04
xtiansimon
1546
answered Aug 8 '16 at 16:30
Valery FradkovValery Fradkov
111
$endgroup$
Logs with different bases cross the line y = 0 at x=1 with different slopes (of the tangent to the curve). Natural base e makes this slope equal to 1.
The number is called e after Leonhard Euler, a mathematician that first gave this number a meaning and found its value. Euler worked on a formula for compounding interest. If r is the annualized interest rate and n is a number of compounding intervals per year, the formula for the amount of investment of $1 after n intervals is: $$(1+frac r n )^n$$ Euler showed that the limit of this value for infinitely large n is $e^r$ where e is $$lim_{ntoinfty} (1+frac 1 n)^n$$ when $n rightarrow infty$. It is approximately 2.718.
edited Nov 21 '16 at 0:04
xtiansimon
1546
answered Aug 8 '16 at 16:30
Valery FradkovValery Fradkov
111
edited Nov 21 '16 at 0:04
xtiansimon
1546
edited Nov 21 '16 at 0:04
xtiansimon
1546
edited Nov 21 '16 at 0:04
xtiansimon
1546
1546
answered Aug 8 '16 at 16:30
Valery FradkovValery Fradkov
111
answered Aug 8 '16 at 16:30
Valery FradkovValery Fradkov
111
answered Aug 8 '16 at 16:30
Valery FradkovValery Fradkov
111
111
add a comment |
add a comment |
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8
$begingroup$
ln denotes the natural log, which has base $e$. Just "log" in this context denotes a log with base $10$. Google "mathematical constant e" for more info.
$endgroup$
– The Chaz 2.0
Dec 11 '11 at 22:20
5
$begingroup$
The formula is ambiguous. Do you mean $$frac{1}{2}ln(x+7)text{ or }frac{1}{2ln(x+7)}$$in the first summand?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:23
2
$begingroup$
@MaX: How do you know which one was meant?
$endgroup$
– Arturo Magidin
Dec 11 '11 at 22:38
7
$begingroup$
The difference is that $ln$ is for children, and $log$ is for grownups.
$endgroup$
– Gerry Myerson
Dec 11 '11 at 22:46
7
$begingroup$
@GerryMyerson : You be nice or I'll whack you across the knuckles with my slide rule.
$endgroup$
– steven gregory
Nov 3 '15 at 18:39