Question

How do you evaluate `log_5 92`?

Answer 1

`approx2.81`

Explanation:

There is a property in logarithms which is `log_a(b)=logb/loga` The proof for this is at the bottom of the answer Using this rule:
`log_5(92)=log92/log5`
Which if you type into a calculator you'll get approximately 2.81.

Proof:
Let `log_ab=x`;
`b=a^x`
`logb=loga^x`
`logb=xloga`
`x=logb/loga`
Therefore `log_ab=logb/loga`

Answer 2

`x=ln(92)/ln(5) ~~2.810` to 3 decimal places

Explanation:

As an example consider `log_10(3) =x`

This mat be written as:`" "10^x=3`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:`" "log_5(92)`

Let `log_5(92)=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(5^x)=ln(92)`

Write this as: `xln(5)=ln(92)`

Divide both sides by `ln(5)` giving:

`x=ln(92)/ln(5) ~~2.810` to 3 decimal places