Taking the logarithm of both sides of $x=(ab/b)^y$

Question

I got confused at $x=(\frac{ab}{b})^y$. If you take the logarithm with base $b$ of $x=(\frac{ab}{b})^y$, wouldn't it be $y\log{bx}=\frac{ab}{b}$. Why did the solution suddenly take the log of $\frac{ab}{b}$? Sorry I'm very confused here.

Answer 1

So from the previous line, we have that $x=a^y$. Now, since $b\neq0$, $\frac{b}{b}=1$ so we may write $$x=a^y=(1\cdot a)^y=\left(\frac{ba}{b}\right)^y.$$ Now, if we take the log base $b$ of both sides, we get $$\log_b{x}=y\log_b{\left(\frac{ba}{b}\right)}=y\left(\log_b{b}+\log_b{a}-\log_b{b}\right)=y\log_b{a}$$

Answer 2

$$x=a^y.\text{ Take note of the fact that }b/b=1,\text{ hence giving }x=a^y=\left(\dfrac{ab}{b}\right)^y.$$