Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$ using logarithms 
Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$.

How do I use logarithms to approach this problem?
 A: Using logarithms:
Since $a^x = b^y$,
$$
  \log a^x = \log b^y \quad \Rightarrow \quad x \log a = y \log b
  \quad \Rightarrow \quad \log a = \frac{y}{x} \log b
$$
Then, since $b^y = (ab)^{xy}$,
$$
  \log b^y = \log (ab)^{xy}
  \quad \Rightarrow \quad
  y \log b = xy \log (ab) = xy \left( \log a + \log b\right)
$$
Let's assume $y \neq 0$ (since if $y=0$, then we must also have $x=0$ and get the required conditions without having $x+y=1$ -- meaning the original question must have had some restriction such as $x, y \neq 0$).  Cancel $y$ on both sides of the last equation:
$$
  \log b = x\left(\log a + \log b\right)
  = x\left( \frac{y}{x} \log b + \log b \right)
  = y \log b + x \log b = (y + x)\log b
$$
Then as long as $b \neq 1$ (which is guaranteed if $x \neq 0$), we know that $\log b \neq 0$, hence we can cancel in the equation:
$$
  \log b = (x+y)\log b
  \quad \Rightarrow \quad
  1 = x + y.
$$
A: How about $x=y=0$ ? Am I missing something?
A: No logarithms are needed:
$$a^x=(ab)^{xy}=a^{xy}b^{xy}=\left(a^x\right)^y\left(b^y\right)^x=\left(a^x\right)^y\left(a^x\right)^x=\left(a^x\right)^{x+y}$$
A: How about this?$$\begin{align} &(ab)^{xy} \\ =& a^{xy}\cdot b^{xy} \\ = & (a^x)^y \cdot (b^y)^x \\ = & (a^x)^y \cdot (a^x)^x \\  = & (a^x)^{x + y} \end{align}$$It suffices to say that $x + y = 1.$
