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
$endgroup$
|
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
$endgroup$
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
$endgroup$
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
edited Jan 21 at 9:42
ss1729
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 $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
$endgroup$
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
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
$endgroup$
$begingroup$
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
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081619%2festimate-the-value-of-log-203%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
$endgroup$
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 $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
$endgroup$
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 $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$
$endgroup$
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
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
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
$endgroup$
$begingroup$
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
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
$endgroup$
$begingroup$
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
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
$endgroup$
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
$$frac4{11}<log_{20}3<frac7{19}$$
$$0.3636364<0.3667258<0.3684211$$
edited Jan 21 at 9:41
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081619%2festimate-the-value-of-log-203%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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