Question

How do you solve `5^x= 20`?

Answer 1

Take logs and use properties of logs to find:

`x = (log 20)/(log 5) = (1+log 2)/(1 - log 2) ~~ 1.86135`

Explanation:

If we take common logs of both sides then we get:

`log 20 = log 5^x = x log 5`

So `x = (log 20)/(log 5)`

We can do a little more with this if we know the log value `log 2 ~~ 0.30103` and that `log 10 = 1`:

`log 5 = log (10/2) = log 10 - log 2 = 1 - log 2`

`~~ 1 - 0.30103 = 0.69897`

`log 20 = log (10*2) = log 10 + log 2 = 1 + log 2`

`~~ 1 + 0.30103 = 1.30103`

So `x = (log 20)/(log 5) ~~ 1.30103 / 0.69897 ~~ 1.86135`