# Question #070b2

Nov 23, 2016

Assuming that the logarithm is base 10:
${\log}_{10} \left(9\right) = \ln \frac{9}{\ln} \left(10\right)$

Aug 18, 2017

${\log}_{10} \left(9\right) = \frac{\ln \left(9\right)}{\ln \left(10\right)}$

#### Explanation:

Consider ${\log}_{10} \left(9\right)$

Set $\text{ } {\log}_{10} \left(9\right) = x$

Another way of writing this is:$\text{ } {10}^{x} = 9$

Set $\text{ } {e}^{b} = 9 \implies \textcolor{red}{b = \ln \left(9\right)}$

But we have:$\text{ } {e}^{b} = 9 = {10}^{x}$

Take natural logs of both sides and using the principle that
$\ln \left({10}^{x}\right)$ is the same as $x \ln \left(10\right)$

$b \ln \left(e\right) = x \ln \left(10\right)$

but $\ln \left(e\right) = 1$

$b = x \ln \left(10\right)$

Thus $x = \frac{\textcolor{red}{b}}{\ln \left(10\right)}$

but $b = \ln \left(9\right)$ giving

$x = \ln \frac{9}{\ln \left(10\right)}$

But we set ${\log}_{10} \left(9\right) = x$ giving:

${\log}_{10} \left(9\right) = \frac{\ln \left(9\right)}{\ln \left(10\right)}$