Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$.
How do I use logarithms to approach this problem?
Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$.
How do I use logarithms to approach this problem?
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 $x + y = 1.$
No logarithms are needed:
$a^x=(ab)^{xy}=a^{xy}b^{xy}=\left(a^x\right)^y\left(b^y\right)^x=\left(a^x\right)^y\left(a^x\right)^x=\left(a^x\right)^{x+y}$
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. $
How about $x=y=0$ ? Am I missing something?