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