$$ log_ax^b=blog_ax $$

if the equation turns to exponential form then…

$a^y=x$

if the equation is to the power of b

$(a^y)^b$=$x^b$

also since

The function of $log_a(x)$ are yielding out the y value or the the exponent value of the inverse original function. With argument being the output value in the original function.

$$ (a^y)^b =(a^{yb})=x^b $$

With those in mind

$$ log_a(x^b)=yb $$

SInce

$$ log_a(x)=y $$

$$ a^{yb}=x^b $$