# How do I calculate the following logarithm?

Say I'd like to calculate the following logarithm:

$$log_{0,1}{\sqrt {10}\over 100}$$

Using the logarithm rules, I do it this way:

$${log_{1\over 10} {\sqrt {10}}} - {log_{1\over 10} {100}}$$

$$={{1\over2}log_{1\over10} {10}} + log_{1\over 10}{10^2}$$

$$={{1\over2}log_{10^{-1}} {10}} + log_{10^{-1}}{10^2}$$

Though, I don't seem to be able to apply the first property of a logarithm:

$$log_aa^c = c$$

$10^{-1}$ is not equal to $10$. How do I calculate the following logarithm?

• $\log_ab=\frac{\log b}{\log a}$ – E.H.E Nov 29 '15 at 10:20

Hint: use the fact that $$10=(10^{-1})^{-1}$$

• Thanks Emilio! I figured it out. – Cesare Nov 29 '15 at 10:19

Using the following rules:

1) $\log_{a}(b)=\frac{\ln(b)}{\ln(a)}$;

2) $\ln\left(\frac{1}{a}\right)=-\ln(a)$;

3) $\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$.

$$\log_{0,1}\left(\frac{\sqrt{10}}{100}\right)=\log_{\frac{1}{10}}\left(\frac{\sqrt{10}}{100}\right)=\frac{\ln\left(\frac{\sqrt{10}}{100}\right)}{\ln\left(\frac{1}{10}\right)}=\frac{\ln\left(\frac{\sqrt{10}}{100}\right)}{-\ln\left(10\right)}=$$ $$-\frac{\ln\left(\sqrt{10}\right)-\ln(100)}{\ln\left(10\right)}=\frac{\ln(100)-\ln\left(\sqrt{10}\right)}{\ln\left(10\right)}=\frac{\ln\left(\frac{100}{\sqrt{10}}\right)}{\ln\left(10\right)}=$$ $$\frac{\ln\left(10\sqrt{10}\right)}{\ln\left(10\right)}=\frac{\ln\left(10^{\frac{3}{2}}\right)}{\ln\left(10\right)}=\frac{\frac{3\ln(10)}{2}}{\ln\left(10\right)}=\frac{3\ln(10)}{2\ln(10)}=\frac{3}{2}\cdot\frac{\ln(10)}{\ln(10)}=\frac{3}{2}$$

Apply the logarithm rule $\color{blue}{\large \log_{a^m}(b^n)=\frac{n}{m}\log_a(b)}$, hence $$\frac{1}{2}\log_{10^{-1}}10-\log_{10^{-1}}10^2$$ $$=-\frac{1}{2}\log_{10}10-2(-1)\log_{10}10$$

$$=-\frac{1}{2}+2=\color{red}{\frac{3}{2}}$$

• Thanks for your help, Harish! – Cesare Nov 29 '15 at 10:29