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

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 ?

• 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? – Satish Ramanathan Jan 21 '19 at 8:03
• Have you copied the problem correctly? – Satish Ramanathan Jan 21 '19 at 8:04
• FWIW, I think the problem should be restated as: Estimate $\log_{20}{3}$. – Jose Arnaldo Bebita-Dris Jan 21 '19 at 8:05
• The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer. – Yves Daoust Jan 21 '19 at 8:14
• 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. – 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 $$\frac{1}{3}<\log_{20}(3)<\frac{1}{2}.$$

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:

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

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

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

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

$$\cdots$$

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

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

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

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

$$\cdots$$

By hand, you could reasonably establish

$$\frac4{11}<\log_{20}3<\frac7{19}$$

$$0.3636364<0.3667258<0.3684211$$

• Nice manipulation ! – Claude Leibovici Jan 21 '19 at 9:06
• @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. – Yves Daoust Jan 21 '19 at 9:08