<s>[INST] 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?
 [/INST] 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:
$$
  \log b = (x+y)\log b
  \quad \Rightarrow \quad
  1 = x + y.
$$
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