Question

How do you condense `1/3log 3x + 2/3log 3x`?

Answer 1

`1/3 log 3x + 2/3 log 3x` `=log(3x)`

Explanation:

`1/3 log 3x + 2/3 log 3x`

`= log(3x)^(1/3) + log(3x)^(2/3)`-> use property `log_bx ^n = nlog _b x`

`=log((3x)^(1/3) (3x)^(2/3))`-> use property `log_b(xy)=log_bx+log_by`

`=log((3x)^(1/3+2/3))`-> use property `x^a*x^b=x^(a+b)`

`=log(3x)`