Question

How do you solve `log_5 6 +log_5 (2x^2)=log_5 48`?

Answer 1

`x=+-2`

Explanation:

Use `log_b m+log_b n = log_b(mn)`

Here,`log_5 6+log_5(2x^2)= log_b (12x^2)=log_5 48`

Log function is a bijective (single valued) function. So,

`12x^2=48`, and so,

`x^2=4 to x =+-2`