Mixing
If $(1 + x)^{10} = 2.5$, what is $x$?
$begingroup$
If
$$(1 + x)^{10} = 2.5$$
, the next step would be
$$10 * log(1 + x) = log(2.5)$$
$$log(1 + x) = frac{log(2.5)}{10}$$
. This is where I am stuck. I know that
$$log_{b}(a + c) = log_{b}(a) + log_bleft( 1 + frac{c}{a} right)$$
, but that does not help me in any way.
logarithms exponentiation
asked Jan 23 at 14:36
NemgathosNemgathos
1374
$endgroup$
add a comment |
$begingroup$
If
$$(1 + x)^{10} = 2.5$$
, the next step would be
$$10 * log(1 + x) = log(2.5)$$
$$log(1 + x) = frac{log(2.5)}{10}$$
. This is where I am stuck. I know that
$$log_{b}(a + c) = log_{b}(a) + log_bleft( 1 + frac{c}{a} right)$$
, but that does not help me in any way.
logarithms exponentiation
asked Jan 23 at 14:36
NemgathosNemgathos
1374
$endgroup$
5
$begingroup$
Just exponentiate both sides to get rid of the $log$ on the left.
$endgroup$
– user3482749
Jan 23 at 14:37
$begingroup$
Alternatively: if $(1+x)^{10} = 2.5$ then $1+x$ is $2.5^{1/10}$. Subtract one.
$endgroup$
– Florian
Jan 23 at 14:41
1
$begingroup$
This isn't quite true. $log (1+x)^{10} = 10cdot log |1+x|$
$endgroup$
– Jakobian
Jan 23 at 14:42
add a comment |
$begingroup$
If
$$(1 + x)^{10} = 2.5$$
, the next step would be
$$10 * log(1 + x) = log(2.5)$$
$$log(1 + x) = frac{log(2.5)}{10}$$
. This is where I am stuck. I know that
$$log_{b}(a + c) = log_{b}(a) + log_bleft( 1 + frac{c}{a} right)$$
, but that does not help me in any way.
logarithms exponentiation
asked Jan 23 at 14:36
NemgathosNemgathos
1374
$endgroup$
If
$$(1 + x)^{10} = 2.5$$
, the next step would be
$$10 * log(1 + x) = log(2.5)$$
$$log(1 + x) = frac{log(2.5)}{10}$$
. This is where I am stuck. I know that
$$log_{b}(a + c) = log_{b}(a) + log_bleft( 1 + frac{c}{a} right)$$
, but that does not help me in any way.
logarithms exponentiation
logarithms exponentiation
asked Jan 23 at 14:36
NemgathosNemgathos
1374
asked Jan 23 at 14:36
NemgathosNemgathos
1374
asked Jan 23 at 14:36
NemgathosNemgathos
1374
asked Jan 23 at 14:36
NemgathosNemgathos
1374
asked Jan 23 at 14:36
NemgathosNemgathos
1374
1374
5
$begingroup$
Just exponentiate both sides to get rid of the $log$ on the left.
$endgroup$
– user3482749
Jan 23 at 14:37
$begingroup$
Alternatively: if $(1+x)^{10} = 2.5$ then $1+x$ is $2.5^{1/10}$. Subtract one.
$endgroup$
– Florian
Jan 23 at 14:41
1
$begingroup$
This isn't quite true. $log (1+x)^{10} = 10cdot log |1+x|$
$endgroup$
– Jakobian
Jan 23 at 14:42
add a comment |
5
$begingroup$
Just exponentiate both sides to get rid of the $log$ on the left.
$endgroup$
– user3482749
Jan 23 at 14:37
$begingroup$
Alternatively: if $(1+x)^{10} = 2.5$ then $1+x$ is $2.5^{1/10}$. Subtract one.
$endgroup$
– Florian
Jan 23 at 14:41
1
$begingroup$
This isn't quite true. $log (1+x)^{10} = 10cdot log |1+x|$
$endgroup$
– Jakobian
Jan 23 at 14:42
5
5
$begingroup$
Just exponentiate both sides to get rid of the $log$ on the left.
$endgroup$
– user3482749
Jan 23 at 14:37
$begingroup$
Just exponentiate both sides to get rid of the $log$ on the left.
$endgroup$
– user3482749
Jan 23 at 14:37
$begingroup$
Alternatively: if $(1+x)^{10} = 2.5$ then $1+x$ is $2.5^{1/10}$. Subtract one.
$endgroup$
– Florian
Jan 23 at 14:41
$begingroup$
Alternatively: if $(1+x)^{10} = 2.5$ then $1+x$ is $2.5^{1/10}$. Subtract one.
$endgroup$
– Florian
Jan 23 at 14:41
1
1
$begingroup$
This isn't quite true. $log (1+x)^{10} = 10cdot log |1+x|$
$endgroup$
– Jakobian
Jan 23 at 14:42
$begingroup$
This isn't quite true. $log (1+x)^{10} = 10cdot log |1+x|$
$endgroup$
– Jakobian
Jan 23 at 14:42
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
When the unknown isn't the exponent, but the base, then logarithms aren't the most natural solution. Roots are:
$$
(1+x)^{10} = 2.5\
1+x = pmsqrt[10]{2.5}\
x = -1pmsqrt[10]{2.5}
$$
edited Jan 23 at 14:46
Jakobian
2,695721
answered Jan 23 at 14:44
ArthurArthur
118k7117200
$endgroup$
$begingroup$
Roots of $2.5$ maybe?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:45
1
$begingroup$
@OscarLanzi Sure. That might be a good idea.
$endgroup$
– Arthur
Jan 23 at 14:46
add a comment |
$begingroup$
Just, $$x=2.5^{0.1}-1$$ or $$x=-2.5^{0.1}-1.$$
answered Jan 23 at 14:43
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$(1+x)^{10} = 2.5$ (original equation)
$1+x = sqrt[10]{2.5} $ (take 10th root of both sides)
$1+x = pm1.09596... $ (simplify)
$x = 0.09596...$ (subtract 1 on both sides, positive root)
or
$x = -2.09596... $ (subtract 1 on both sides, negative root)
edited Jan 28 at 23:13
Marvin Cohen
160117
answered Jan 28 at 4:13
DhanushDhanush
212
$endgroup$
add a comment |
$begingroup$
If the equation $log(1+x) = frac{log(2.5)}{10}$ holds then this equation holds no matter what base the logs were taken in (as long as the logs on both sides were taken to same base). Make sure you see why.
Calculate $a=log_e(2.5)$. Then $1+x = e^{a/10}$ which gives $x= e^{a/10}-1$.
If you are more comfortable working base 10, then
$a=log_{10}(2.5)$. Then $1+x = 10^{a/10}$ which gives $x= 10^{a/10}-1$.
ETA: you should check that this answer and Arthur's above give the same result.
answered Jan 23 at 14:44
MikeMike
4,386412
$endgroup$
2
$begingroup$
Taking logarithms is ok, but arriving at $log(1+x) = frac{log(2.5)}{10}$ is not. We drop a solution by saying that $log(1+x)^{10} = 10log(1+x)$ instead of $log(1+x)^{10} = 10log|x+1|$
$endgroup$
– Jakobian
Jan 23 at 14:49
$begingroup$
You are correct @Jakobian
$endgroup$
– Mike
Jan 23 at 15:11
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When the unknown isn't the exponent, but the base, then logarithms aren't the most natural solution. Roots are:
$$
(1+x)^{10} = 2.5\
1+x = pmsqrt[10]{2.5}\
x = -1pmsqrt[10]{2.5}
$$
edited Jan 23 at 14:46
Jakobian
2,695721
answered Jan 23 at 14:44
ArthurArthur
118k7117200
$endgroup$
$begingroup$
Roots of $2.5$ maybe?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:45
1
$begingroup$
@OscarLanzi Sure. That might be a good idea.
$endgroup$
– Arthur
Jan 23 at 14:46
add a comment |
$begingroup$
When the unknown isn't the exponent, but the base, then logarithms aren't the most natural solution. Roots are:
$$
(1+x)^{10} = 2.5\
1+x = pmsqrt[10]{2.5}\
x = -1pmsqrt[10]{2.5}
$$
edited Jan 23 at 14:46
Jakobian
2,695721
answered Jan 23 at 14:44
ArthurArthur
118k7117200
$endgroup$
$begingroup$
Roots of $2.5$ maybe?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:45
1
$begingroup$
@OscarLanzi Sure. That might be a good idea.
$endgroup$
– Arthur
Jan 23 at 14:46
add a comment |
$begingroup$
When the unknown isn't the exponent, but the base, then logarithms aren't the most natural solution. Roots are:
$$
(1+x)^{10} = 2.5\
1+x = pmsqrt[10]{2.5}\
x = -1pmsqrt[10]{2.5}
$$
edited Jan 23 at 14:46
Jakobian
2,695721
answered Jan 23 at 14:44
ArthurArthur
118k7117200
$endgroup$
When the unknown isn't the exponent, but the base, then logarithms aren't the most natural solution. Roots are:
$$
(1+x)^{10} = 2.5\
1+x = pmsqrt[10]{2.5}\
x = -1pmsqrt[10]{2.5}
$$
edited Jan 23 at 14:46
Jakobian
2,695721
answered Jan 23 at 14:44
ArthurArthur
118k7117200
edited Jan 23 at 14:46
Jakobian
2,695721
edited Jan 23 at 14:46
Jakobian
2,695721
edited Jan 23 at 14:46
Jakobian
2,695721
2,695721
answered Jan 23 at 14:44
ArthurArthur
118k7117200
answered Jan 23 at 14:44
ArthurArthur
118k7117200
answered Jan 23 at 14:44
ArthurArthur
118k7117200
118k7117200
$begingroup$
Roots of $2.5$ maybe?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:45
1
$begingroup$
@OscarLanzi Sure. That might be a good idea.
$endgroup$
– Arthur
Jan 23 at 14:46
add a comment |
$begingroup$
Roots of $2.5$ maybe?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:45
1
$begingroup$
@OscarLanzi Sure. That might be a good idea.
$endgroup$
– Arthur
Jan 23 at 14:46
$begingroup$
Roots of $2.5$ maybe?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:45
$begingroup$
Roots of $2.5$ maybe?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:45
1
1
$begingroup$
@OscarLanzi Sure. That might be a good idea.
$endgroup$
– Arthur
Jan 23 at 14:46
$begingroup$
@OscarLanzi Sure. That might be a good idea.
$endgroup$
– Arthur
Jan 23 at 14:46
add a comment |
$begingroup$
Just, $$x=2.5^{0.1}-1$$ or $$x=-2.5^{0.1}-1.$$
answered Jan 23 at 14:43
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Just, $$x=2.5^{0.1}-1$$ or $$x=-2.5^{0.1}-1.$$
answered Jan 23 at 14:43
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Just, $$x=2.5^{0.1}-1$$ or $$x=-2.5^{0.1}-1.$$
answered Jan 23 at 14:43
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
Just, $$x=2.5^{0.1}-1$$ or $$x=-2.5^{0.1}-1.$$
answered Jan 23 at 14:43
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 14:43
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 14:43
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 14:43
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
$begingroup$
$(1+x)^{10} = 2.5$ (original equation)
$1+x = sqrt[10]{2.5} $ (take 10th root of both sides)
$1+x = pm1.09596... $ (simplify)
$x = 0.09596...$ (subtract 1 on both sides, positive root)
or
$x = -2.09596... $ (subtract 1 on both sides, negative root)
edited Jan 28 at 23:13
Marvin Cohen
160117
answered Jan 28 at 4:13
DhanushDhanush
212
$endgroup$
add a comment |
$begingroup$
$(1+x)^{10} = 2.5$ (original equation)
$1+x = sqrt[10]{2.5} $ (take 10th root of both sides)
$1+x = pm1.09596... $ (simplify)
$x = 0.09596...$ (subtract 1 on both sides, positive root)
or
$x = -2.09596... $ (subtract 1 on both sides, negative root)
edited Jan 28 at 23:13
Marvin Cohen
160117
answered Jan 28 at 4:13
DhanushDhanush
212
$endgroup$
add a comment |
$begingroup$
$(1+x)^{10} = 2.5$ (original equation)
$1+x = sqrt[10]{2.5} $ (take 10th root of both sides)
$1+x = pm1.09596... $ (simplify)
$x = 0.09596...$ (subtract 1 on both sides, positive root)
or
$x = -2.09596... $ (subtract 1 on both sides, negative root)
edited Jan 28 at 23:13
Marvin Cohen
160117
answered Jan 28 at 4:13
DhanushDhanush
212
$endgroup$
$(1+x)^{10} = 2.5$ (original equation)
$1+x = sqrt[10]{2.5} $ (take 10th root of both sides)
$1+x = pm1.09596... $ (simplify)
$x = 0.09596...$ (subtract 1 on both sides, positive root)
or
$x = -2.09596... $ (subtract 1 on both sides, negative root)
edited Jan 28 at 23:13
Marvin Cohen
160117
answered Jan 28 at 4:13
DhanushDhanush
212
edited Jan 28 at 23:13
Marvin Cohen
160117
edited Jan 28 at 23:13
Marvin Cohen
160117
edited Jan 28 at 23:13
Marvin Cohen
160117
160117
answered Jan 28 at 4:13
DhanushDhanush
212
answered Jan 28 at 4:13
DhanushDhanush
212
answered Jan 28 at 4:13
DhanushDhanush
212
212
add a comment |
add a comment |
$begingroup$
If the equation $log(1+x) = frac{log(2.5)}{10}$ holds then this equation holds no matter what base the logs were taken in (as long as the logs on both sides were taken to same base). Make sure you see why.
Calculate $a=log_e(2.5)$. Then $1+x = e^{a/10}$ which gives $x= e^{a/10}-1$.
If you are more comfortable working base 10, then
$a=log_{10}(2.5)$. Then $1+x = 10^{a/10}$ which gives $x= 10^{a/10}-1$.
ETA: you should check that this answer and Arthur's above give the same result.
answered Jan 23 at 14:44
MikeMike
4,386412
$endgroup$
2
$begingroup$
Taking logarithms is ok, but arriving at $log(1+x) = frac{log(2.5)}{10}$ is not. We drop a solution by saying that $log(1+x)^{10} = 10log(1+x)$ instead of $log(1+x)^{10} = 10log|x+1|$
$endgroup$
– Jakobian
Jan 23 at 14:49
$begingroup$
You are correct @Jakobian
$endgroup$
– Mike
Jan 23 at 15:11
add a comment |
$begingroup$
If the equation $log(1+x) = frac{log(2.5)}{10}$ holds then this equation holds no matter what base the logs were taken in (as long as the logs on both sides were taken to same base). Make sure you see why.
Calculate $a=log_e(2.5)$. Then $1+x = e^{a/10}$ which gives $x= e^{a/10}-1$.
If you are more comfortable working base 10, then
$a=log_{10}(2.5)$. Then $1+x = 10^{a/10}$ which gives $x= 10^{a/10}-1$.
ETA: you should check that this answer and Arthur's above give the same result.
answered Jan 23 at 14:44
MikeMike
4,386412
$endgroup$
2
$begingroup$
Taking logarithms is ok, but arriving at $log(1+x) = frac{log(2.5)}{10}$ is not. We drop a solution by saying that $log(1+x)^{10} = 10log(1+x)$ instead of $log(1+x)^{10} = 10log|x+1|$
$endgroup$
– Jakobian
Jan 23 at 14:49
$begingroup$
You are correct @Jakobian
$endgroup$
– Mike
Jan 23 at 15:11
add a comment |
$begingroup$
If the equation $log(1+x) = frac{log(2.5)}{10}$ holds then this equation holds no matter what base the logs were taken in (as long as the logs on both sides were taken to same base). Make sure you see why.
Calculate $a=log_e(2.5)$. Then $1+x = e^{a/10}$ which gives $x= e^{a/10}-1$.
If you are more comfortable working base 10, then
$a=log_{10}(2.5)$. Then $1+x = 10^{a/10}$ which gives $x= 10^{a/10}-1$.
ETA: you should check that this answer and Arthur's above give the same result.
answered Jan 23 at 14:44
MikeMike
4,386412
$endgroup$
If the equation $log(1+x) = frac{log(2.5)}{10}$ holds then this equation holds no matter what base the logs were taken in (as long as the logs on both sides were taken to same base). Make sure you see why.
Calculate $a=log_e(2.5)$. Then $1+x = e^{a/10}$ which gives $x= e^{a/10}-1$.
If you are more comfortable working base 10, then
$a=log_{10}(2.5)$. Then $1+x = 10^{a/10}$ which gives $x= 10^{a/10}-1$.
ETA: you should check that this answer and Arthur's above give the same result.
answered Jan 23 at 14:44
MikeMike
4,386412
answered Jan 23 at 14:44
MikeMike
4,386412
answered Jan 23 at 14:44
MikeMike
4,386412
answered Jan 23 at 14:44
MikeMike
4,386412
4,386412
2
$begingroup$
Taking logarithms is ok, but arriving at $log(1+x) = frac{log(2.5)}{10}$ is not. We drop a solution by saying that $log(1+x)^{10} = 10log(1+x)$ instead of $log(1+x)^{10} = 10log|x+1|$
$endgroup$
– Jakobian
Jan 23 at 14:49
$begingroup$
You are correct @Jakobian
$endgroup$
– Mike
Jan 23 at 15:11
add a comment |
2
$begingroup$
Taking logarithms is ok, but arriving at $log(1+x) = frac{log(2.5)}{10}$ is not. We drop a solution by saying that $log(1+x)^{10} = 10log(1+x)$ instead of $log(1+x)^{10} = 10log|x+1|$
$endgroup$
– Jakobian
Jan 23 at 14:49
$begingroup$
You are correct @Jakobian
$endgroup$
– Mike
Jan 23 at 15:11
2
2
$begingroup$
Taking logarithms is ok, but arriving at $log(1+x) = frac{log(2.5)}{10}$ is not. We drop a solution by saying that $log(1+x)^{10} = 10log(1+x)$ instead of $log(1+x)^{10} = 10log|x+1|$
$endgroup$
– Jakobian
Jan 23 at 14:49
$begingroup$
Taking logarithms is ok, but arriving at $log(1+x) = frac{log(2.5)}{10}$ is not. We drop a solution by saying that $log(1+x)^{10} = 10log(1+x)$ instead of $log(1+x)^{10} = 10log|x+1|$
$endgroup$
– Jakobian
Jan 23 at 14:49
$begingroup$
You are correct @Jakobian
$endgroup$
– Mike
Jan 23 at 15:11
$begingroup$
You are correct @Jakobian
$endgroup$
– Mike
Jan 23 at 15:11
add a comment |
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5
$begingroup$
Just exponentiate both sides to get rid of the $log$ on the left.
$endgroup$
– user3482749
Jan 23 at 14:37
$begingroup$
Alternatively: if $(1+x)^{10} = 2.5$ then $1+x$ is $2.5^{1/10}$. Subtract one.
$endgroup$
– Florian
Jan 23 at 14:41
1
$begingroup$
This isn't quite true. $log (1+x)^{10} = 10cdot log |1+x|$
$endgroup$
– Jakobian
Jan 23 at 14:42