Answer
See proof
Work Step by Step
We are given the properties:
$b^{x+y}=b^xb^y$
$log_b (xy)=log_b x+log_b y$
Let's supose $b^{x+y}=b^xb^y$ is true and let's note:
$A=\log_b (xy)$
$B=\log_b x$
$C=\log_b y$.
We have:
$A=\log_b (xy)\Rightarrow xy=b^A$
$B=\log_b x\Rightarrow x=b^B$
$C=\log_b y\Rightarrow y=b^C$.
$b^A=xy=b^B\cdot b^C$ (1)
We are given: $b^{x+y}=b^xb^y$. We apply it in (1):
$b^{B+C}=b^B\cdot b^C$ (2)
From (1) and (2) we get:
$b^A=b^{B+C}$
Substitute the expressions of $A,B,C$:
$b^{\log_b (xy)}=b^{\log_b x}\cdot b^{\log_b y}$
The exponents are equal as the bases are:
$log_b (xy)=log_b x+log_b y$
