# How do you evaluate log_5 92?

Apr 21, 2016

$\approx 2.81$

#### Explanation:

There is a property in logarithms which is ${\log}_{a} \left(b\right) = \log \frac{b}{\log} a$ The proof for this is at the bottom of the answer Using this rule:
${\log}_{5} \left(92\right) = \log \frac{92}{\log} 5$
Which if you type into a calculator you'll get approximately 2.81.

Proof:
Let ${\log}_{a} b = x$;
$b = {a}^{x}$
$\log b = \log {a}^{x}$
$\log b = x \log a$
$x = \log \frac{b}{\log} a$
Therefore ${\log}_{a} b = \log \frac{b}{\log} a$

Apr 21, 2016

$x = \ln \frac{92}{\ln} \left(5\right) \approx 2.810$ to 3 decimal places

#### Explanation:

As an example consider ${\log}_{10} \left(3\right) = x$

This mat be written as:$\text{ } {10}^{x} = 3$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:$\text{ } {\log}_{5} \left(92\right)$

Let ${\log}_{5} \left(92\right) = x$

The we have: ${5}^{x} = 92$

You can use log base 10 or natural loges (ln). This will work for either.

Take logs of both sides

$\ln \left({5}^{x}\right) = \ln \left(92\right)$

Write this as: $x \ln \left(5\right) = \ln \left(92\right)$

Divide both sides by $\ln \left(5\right)$ giving:

$x = \ln \frac{92}{\ln} \left(5\right) \approx 2.810$ to 3 decimal places