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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?

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up vote 8 down vote accepted

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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No logarithms are needed:


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Hm, I wouldn't be surprised if I'm missing something obvious here, but how do you get $(a^x)^1 = (a^x)^{x+y}$ implies $1 = x+y$ without logarithms? In other words, how can we get this result without implicitly taking a logarithm while stating that $x \to a^x$ is injective? Also, I think the case $a = b = 1$ has to be excluded. – user38355 Dec 14 '12 at 1:14
@brom: Depends on the order in which one defines and proves things about exponentials and logs. It’s quite possible to define the exponential functions and prove that they’re injective (for base not equal $1$, of course) before dealing with logs at all. – Brian M. Scott Dec 14 '12 at 1:19
Indeed $a=b=1$ has to be excluded. $f(x)=a^x$ is injective provided $a > 0$ and $a \neq 1$ because this function is monotone increasing for $a > 1$ and monotone decreasing for $a < 1$. – Gyu Eun Lee Dec 14 '12 at 1:20
@BrianM.Scott What about Hugo's response? – Alan Dec 14 '12 at 4:19
@Alan: What about it? It’s perfectly correct. Mine was never intended to be a complete answer: I was addressing the main case, presumably the one of greatest interest to the OP. – Brian M. Scott Dec 14 '12 at 4:46

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 integers – Henry Dec 14 '12 at 7:21

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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