Question

How do you solve `log_10 (2^x)=log_10 32`?

Answer 1

`x=5`

Explanation:

Rewrite `32` as a productof `2`

`32=2*2*2*2*2=2^5`

`:.log_10(2^x)=log_10(2^5)`

`2^x=2^5`

`x=5`