0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ $\log_ab=\frac{\log b}{\log a}$ $\endgroup$ – E.H.E Nov 29 '15 at 10:20
2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Thanks Emilio! I figured it out. $\endgroup$ – Cesare Nov 29 '15 at 10:19
2
$\begingroup$

Using the following rules:

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

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

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


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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ Thanks for your help, Harish! $\endgroup$ – Cesare Nov 29 '15 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.