Mixing
Estimate the value of $log_{20}3$
$begingroup$
Estimate the value of $log_{20}3$
My Attempt
$$
y=log_{20}3=frac{1}{log_{3}20}=frac{log_{c}3}{log_c20}=frac{log_c3}{log_c5+log_c4}\
=frac{1}{frac{log_c5}{log_c3}+frac{log_c4}{log_c3}}=frac{1}{log_35+log_34}\
x=log_35implies3^x=5implies x<2;&;x>1\
z=log_34implies3^z=4implies z<2;&;z>1\
x+z<4;&;x+z>2\
yinBig(frac{1}{4},frac{1}{2}Big)
$$
My reference gives the solution $Big(frac{1}{3},frac{1}{2}Big)$, it seems $frac{1}{4}$ is not the lowest limit of $log_{20}3$, what's the easiest way to calculate it ?
number-theory logarithms
asked Jan 21 at 7:54
ss1729ss1729
2,00111024
$endgroup$
|
show 1 more comment
$begingroup$
Estimate the value of $log_{20}3$
My Attempt
$$
y=log_{20}3=frac{1}{log_{3}20}=frac{log_{c}3}{log_c20}=frac{log_c3}{log_c5+log_c4}\
=frac{1}{frac{log_c5}{log_c3}+frac{log_c4}{log_c3}}=frac{1}{log_35+log_34}\
x=log_35implies3^x=5implies x<2;&;x>1\
z=log_34implies3^z=4implies z<2;&;z>1\
x+z<4;&;x+z>2\
yinBig(frac{1}{4},frac{1}{2}Big)
$$
My reference gives the solution $Big(frac{1}{3},frac{1}{2}Big)$, it seems $frac{1}{4}$ is not the lowest limit of $log_{20}3$, what's the easiest way to calculate it ?
number-theory logarithms
asked Jan 21 at 7:54
ss1729ss1729
2,00111024
$endgroup$
1
$begingroup$
y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:03
1
$begingroup$
Have you copied the problem correctly?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:04
7
$begingroup$
FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 21 at 8:05
2
$begingroup$
The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
$endgroup$
– Yves Daoust
Jan 21 at 8:14
1
$begingroup$
Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
$endgroup$
– zwim
Jan 21 at 8:58
|
show 1 more comment
$begingroup$
Estimate the value of $log_{20}3$
My Attempt
$$
y=log_{20}3=frac{1}{log_{3}20}=frac{log_{c}3}{log_c20}=frac{log_c3}{log_c5+log_c4}\
=frac{1}{frac{log_c5}{log_c3}+frac{log_c4}{log_c3}}=frac{1}{log_35+log_34}\
x=log_35implies3^x=5implies x<2;&;x>1\
z=log_34implies3^z=4implies z<2;&;z>1\
x+z<4;&;x+z>2\
yinBig(frac{1}{4},frac{1}{2}Big)
$$
My reference gives the solution $Big(frac{1}{3},frac{1}{2}Big)$, it seems $frac{1}{4}$ is not the lowest limit of $log_{20}3$, what's the easiest way to calculate it ?
number-theory logarithms
asked Jan 21 at 7:54
ss1729ss1729
2,00111024
$endgroup$
Estimate the value of $log_{20}3$
My Attempt
$$
y=log_{20}3=frac{1}{log_{3}20}=frac{log_{c}3}{log_c20}=frac{log_c3}{log_c5+log_c4}\
=frac{1}{frac{log_c5}{log_c3}+frac{log_c4}{log_c3}}=frac{1}{log_35+log_34}\
x=log_35implies3^x=5implies x<2;&;x>1\
z=log_34implies3^z=4implies z<2;&;z>1\
x+z<4;&;x+z>2\
yinBig(frac{1}{4},frac{1}{2}Big)
$$
My reference gives the solution $Big(frac{1}{3},frac{1}{2}Big)$, it seems $frac{1}{4}$ is not the lowest limit of $log_{20}3$, what's the easiest way to calculate it ?
number-theory logarithms
number-theory logarithms
asked Jan 21 at 7:54
ss1729ss1729
2,00111024
asked Jan 21 at 7:54
ss1729ss1729
2,00111024
edited Jan 21 at 9:42
ss1729
asked Jan 21 at 7:54
ss1729ss1729
2,00111024
asked Jan 21 at 7:54
ss1729ss1729
2,00111024
asked Jan 21 at 7:54
ss1729ss1729
2,00111024
2,00111024
1
$begingroup$
y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:03
1
$begingroup$
Have you copied the problem correctly?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:04
7
$begingroup$
FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 21 at 8:05
2
$begingroup$
The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
$endgroup$
– Yves Daoust
Jan 21 at 8:14
1
$begingroup$
Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
$endgroup$
– zwim
Jan 21 at 8:58
|
show 1 more comment
1
$begingroup$
y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:03
1
$begingroup$
Have you copied the problem correctly?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:04
7
$begingroup$
FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 21 at 8:05
2
$begingroup$
The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
$endgroup$
– Yves Daoust
Jan 21 at 8:14
1
$begingroup$
Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
$endgroup$
– zwim
Jan 21 at 8:58
1
1
$begingroup$
y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:03
$begingroup$
y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:03
1
1
$begingroup$
Have you copied the problem correctly?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:04
$begingroup$
Have you copied the problem correctly?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:04
7
7
$begingroup$
FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 21 at 8:05
$begingroup$
FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 21 at 8:05
2
2
$begingroup$
The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
$endgroup$
– Yves Daoust
Jan 21 at 8:14
$begingroup$
The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
$endgroup$
– Yves Daoust
Jan 21 at 8:14
1
1
$begingroup$
Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
$endgroup$
– zwim
Jan 21 at 8:58
$begingroup$
Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
$endgroup$
– zwim
Jan 21 at 8:58
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Estimation of $log_{20}(3)$
Using $$9<20<27.$$
$$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
So $$2<log_{3}(20)<3$$
So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
answered Jan 21 at 8:08
DXTDXT
5,9742732
$endgroup$
add a comment |
$begingroup$
Write the sequence of powers of $3$ and $20$:
$$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$
$$1,20,400,8000,160000,3200000,64000000,cdots$$
This gives you many rational upper and lower bounds:
$$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$
$$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$
$$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$
$$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$
$$cdots$$
$$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$
$$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$
$$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$
$$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$
$$cdots$$
By hand, you could reasonably establish
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
answered Jan 21 at 8:31
Yves DaoustYves Daoust
129k676227
$endgroup$
$begingroup$
Nice manipulation !
$endgroup$
– Claude Leibovici
Jan 21 at 9:06
$begingroup$
@ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
$endgroup$
– Yves Daoust
Jan 21 at 9:08
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
Estimation of $log_{20}(3)$
Using $$9<20<27.$$
$$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
So $$2<log_{3}(20)<3$$
So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
answered Jan 21 at 8:08
DXTDXT
5,9742732
$endgroup$
add a comment |
$begingroup$
Estimation of $log_{20}(3)$
Using $$9<20<27.$$
$$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
So $$2<log_{3}(20)<3$$
So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
answered Jan 21 at 8:08
DXTDXT
5,9742732
$endgroup$
add a comment |
$begingroup$
Estimation of $log_{20}(3)$
Using $$9<20<27.$$
$$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
So $$2<log_{3}(20)<3$$
So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
answered Jan 21 at 8:08
DXTDXT
5,9742732
$endgroup$
Estimation of $log_{20}(3)$
Using $$9<20<27.$$
$$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
So $$2<log_{3}(20)<3$$
So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
answered Jan 21 at 8:08
DXTDXT
5,9742732
answered Jan 21 at 8:08
DXTDXT
5,9742732
answered Jan 21 at 8:08
DXTDXT
5,9742732
answered Jan 21 at 8:08
DXTDXT
5,9742732
5,9742732
add a comment |
add a comment |
$begingroup$
Write the sequence of powers of $3$ and $20$:
$$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$
$$1,20,400,8000,160000,3200000,64000000,cdots$$
This gives you many rational upper and lower bounds:
$$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$
$$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$
$$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$
$$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$
$$cdots$$
$$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$
$$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$
$$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$
$$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$
$$cdots$$
By hand, you could reasonably establish
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
answered Jan 21 at 8:31
Yves DaoustYves Daoust
129k676227
$endgroup$
$begingroup$
Nice manipulation !
$endgroup$
– Claude Leibovici
Jan 21 at 9:06
$begingroup$
@ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
$endgroup$
– Yves Daoust
Jan 21 at 9:08
add a comment |
$begingroup$
Write the sequence of powers of $3$ and $20$:
$$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$
$$1,20,400,8000,160000,3200000,64000000,cdots$$
This gives you many rational upper and lower bounds:
$$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$
$$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$
$$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$
$$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$
$$cdots$$
$$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$
$$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$
$$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$
$$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$
$$cdots$$
By hand, you could reasonably establish
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
answered Jan 21 at 8:31
Yves DaoustYves Daoust
129k676227
$endgroup$
$begingroup$
Nice manipulation !
$endgroup$
– Claude Leibovici
Jan 21 at 9:06
$begingroup$
@ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
$endgroup$
– Yves Daoust
Jan 21 at 9:08
add a comment |
$begingroup$
Write the sequence of powers of $3$ and $20$:
$$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$
$$1,20,400,8000,160000,3200000,64000000,cdots$$
This gives you many rational upper and lower bounds:
$$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$
$$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$
$$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$
$$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$
$$cdots$$
$$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$
$$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$
$$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$
$$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$
$$cdots$$
By hand, you could reasonably establish
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
answered Jan 21 at 8:31
Yves DaoustYves Daoust
129k676227
$endgroup$
Write the sequence of powers of $3$ and $20$:
$$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$
$$1,20,400,8000,160000,3200000,64000000,cdots$$
This gives you many rational upper and lower bounds:
$$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$
$$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$
$$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$
$$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$
$$cdots$$
$$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$
$$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$
$$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$
$$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$
$$cdots$$
By hand, you could reasonably establish
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
answered Jan 21 at 8:31
Yves DaoustYves Daoust
129k676227
edited Jan 21 at 9:41
answered Jan 21 at 8:31
Yves DaoustYves Daoust
129k676227
answered Jan 21 at 8:31
Yves DaoustYves Daoust
129k676227
answered Jan 21 at 8:31
Yves DaoustYves Daoust
129k676227
129k676227
$begingroup$
Nice manipulation !
$endgroup$
– Claude Leibovici
Jan 21 at 9:06
$begingroup$
@ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
$endgroup$
– Yves Daoust
Jan 21 at 9:08
add a comment |
$begingroup$
Nice manipulation !
$endgroup$
– Claude Leibovici
Jan 21 at 9:06
$begingroup$
@ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
$endgroup$
– Yves Daoust
Jan 21 at 9:08
$begingroup$
Nice manipulation !
$endgroup$
– Claude Leibovici
Jan 21 at 9:06
$begingroup$
Nice manipulation !
$endgroup$
– Claude Leibovici
Jan 21 at 9:06
$begingroup$
@ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
$endgroup$
– Yves Daoust
Jan 21 at 9:08
$begingroup$
@ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
$endgroup$
– Yves Daoust
Jan 21 at 9:08
add a comment |
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$begingroup$
y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:03
1
$begingroup$
Have you copied the problem correctly?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:04
7
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FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
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– Jose Arnaldo Bebita Dris
Jan 21 at 8:05
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The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
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– Yves Daoust
Jan 21 at 8:14
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Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
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– zwim
Jan 21 at 8:58