For $a,b > 1$ and $y,z > 0$, show by using only the power rules and the definition $x = \log_b(y) \Leftrightarrow_{df} y = b^x$ that $\log_b{(y\cdot z)} = \log_b(y)+log_b(z)$.
I don't get how to solve this without using $b^{log_b{x}} = x$, which isn't a power rule. Is there another way, maybe by using substitution? My "solution" using the mentioned rule:
$\log_b{(y\cdot z)} = \log_b{(y)}+ \log_b{(z)} \Leftrightarrow_{df}$
$b^{\log_b{(y)}+\log_b{(z)}} = y \cdot z \Leftrightarrow_{Power Rule}$
$b^{\log_b{(y)}} \cdot b^{\log_b{(z)}} = y \cdot z \Leftrightarrow_{Forbidden Rule}$
$y \cdot z = y \cdot z$
By definition $\log_b xy$ is that number $N$ such that $b^N = xy$. Likewise, $\log_b x=M$ is such that $b^M = x$ and $\log_b y=L$ is such that $b^L = y$. Hence $$b^N = xy = b^Mb^L=b^{M+L}$$ and therefore $N=M+L$, which means $\log_b xy=\log_b x+\log_b y$.