Question

How do you calculate `log_5(4)`?

Answer 1

`log_5(4)=0.8614`

Explanation:

Let `log_ba=x`, then `b^x=a`.

If `a=10^n` and `b=10^m`, then `n=loga` and `m=logb` and

`b^x=a` becomes `(10^m)^x=10^n` or `10^(mx)=10^n` i.e.

`mx=n`

Hence `x=n/m=loga/logb`

Thus `log_5(4)=log4/log5=0.6021/0.6990=0.8614`