Estimate the value of $\log_{20}3$

My Attempt $$ y=\log_{20}3=\frac{1}{\log_{3}20}=\frac{\log_{c}3}{\log_c20}=\frac{\log_c3}{\log_c5+\log_c4}\\ =\frac{1}{\frac{\log_c5}{\log_c3}+\frac{\log_c4}{\log_c3}}=\frac{1}{\log_35+\log_34}\\ x=\log_35\implies3^x=5\implies x<2\;\&\;x>1\\ z=\log_34\implies3^z=4\implies z<2\;\&\;z>1\\ x+z<4\;\&\;x+z>2\\ y\in\Big(\frac{1}{4},\frac{1}{2}\Big) $$ 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 ?

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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 '19 at 8:03
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    $\begingroup$ Have you copied the problem correctly? $\endgroup$ – Satish Ramanathan Jan 21 '19 at 8:04
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    $\begingroup$ FWIW, I think the problem should be restated as: Estimate $\log_{20}{3}$. $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 21 '19 at 8:05
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    $\begingroup$ The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer. $\endgroup$ – Yves Daoust Jan 21 '19 at 8:14
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    $\begingroup$ 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 '19 at 8:58

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 $$\frac{1}{3}<\log_{20}(3)<\frac{1}{2}.$$

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Write the sequence of powers of $3$ and $20$:



This gives you many rational upper and lower bounds:

$$3^2<20^1\implies 2\log3<1\log20\implies\log_{20}3<\frac12$$

$$3^5<20^2\implies 5\log3<2\log20\implies\log_{20}3<\frac25$$

$$3^8<20^3\implies 8\log3<3\log20\implies\log_{20}3<\frac38$$

$$3^{10}<20^4\implies 10\log3<4\log20\implies\log_{20}3<\frac25$$


$$3^3>20^1\implies 3\log3>1\log20\implies\log_{20}3>\frac13$$

$$3^6>20^2\implies 6\log3>2\log20\implies\log_{20}3>\frac13$$

$$3^9>20^3\implies 9\log3>3\log20\implies\log_{20}3>\frac13$$

$$3^{11}>20^4\implies 11\log3>4\log20\implies\log_{20}3>\frac4{11}$$


By hand, you could reasonably establish



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  • $\begingroup$ Nice manipulation ! $\endgroup$ – Claude Leibovici Jan 21 '19 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 '19 at 9:08

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