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

How do I use logarithms to approach this problem?

4 Answers 4

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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 $x + y = 1.$

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No logarithms are needed:

$a^x=(ab)^{xy}=a^{xy}b^{xy}=\left(a^x\right)^y\left(b^y\right)^x=\left(a^x\right)^y\left(a^x\right)^x=\left(a^x\right)^{x+y}$

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    @Assad: Let $c=\log_a b$. If you take logs base $a$, you get $x=yc=xy\log_a(ab)=xy(1+c)=xy+xyc\;.$ Since $yc=x$, $xyc=x^2$, and we have $x=xy+x^2$. Assuming that $x\ne0$, we can divide through by $x$ to get $1=y+x$.2012-12-14
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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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How about $x=y=0$ ? Am I missing something?

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    You missed $a=b=1$, or $a=b=0$, or $a=b=-1$ with $x$ and $y$ both even integers2012-12-14