How do you evaluate log_5 92?

2 Answers
Apr 21, 2016

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

Apr 21, 2016

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