Question

How do you solve `0.38 = log x` ?

Answer 1

`x= 10^0.38 approx 2.39883292`
[Assuming `log = log_10`]

Explanation:

`0.38 = log x`

As rightly pointed out in the comment below, this can only be solved if we are told the base of the log. Since, this question is in the "Common Logs" section, I'm going to assume that this is `log_10`

Hence, we have: `0.38 = log_10 x`

`log_10x = 0.38`

`x= 10^0.38`

`x approx 2.39883292`