Question

How do you solve `\log_8 2 + \log _ { 8} 2x ^ { 2} = \log _ { 8} 64`?

Answer 1

`x=4` and `x=-4`

Explanation:

`log_a(b)+log_a(c)=log_a(bc)`

`log_(8)2+log_(8)2x^2=log_(8)64`

`log_(8)4x^2=log_(8)64=>4x^2=64`

`4x^2=64`

`x^2=64/4=16`

`x=sqrt(16)=> x=+-4`