Question

How do you solve `log4x - log16 = 1`?

Answer 1

Use the log rule `log_an - log_am = log_a(n/m)`

Explanation:

`log_10((4x)/16) = 1`

`(4x)/16 = 10^1`

`4x = 10(16)`

`4x = 160`

`x = 40`

Hopefully this helps!