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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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    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.2012-12-14
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    @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.2012-12-14
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    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 $a < 1$.2012-12-14
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    @BrianM.Scott What about Hugo's response?2012-12-14
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    @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.2012-12-14
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    @BrianM.Scott Well, I'm not a mathematician but I guess it seems to me that you can't prove that x+y=1 since there's other answers that work? Maybe he left out some part of the problem?2012-12-14
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    @Alan: I don’t know whether the original version of the problem included the necessary minor side conditions or not. I thought it pretty obvious that the intent was to exclude the special cases $x=y=0$ and $a=b=1$, so I didn’t comment on them.2012-12-14
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    @BrianM.Scott Alright thanks, I think a=b=0 is another as well2012-12-14
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    Yet another would be $a=b=-1$ and $x$ and $y$ any even integers2012-12-14
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    @Henry: In the context of exponentials with things called $x$ and $y$ as exponents I really think that it’s safe to assume that $a$ and $b$ are intended to be non-negative.2012-12-14
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    @BrianM.Scott: You have shown $a^x=\left(a^x\right)^{x+y}$. The next step is to say either $a^x=0$ or $a^{x(x+y-1)}=1$ and draw conclusions from that, which could include (i) $a=0$ or (ii) $a=1$ or (iii) $x(x+y-1)=0$ or (iv) $a=-1$ and $x(x+y-1)$ is even. Then for each of these look at the consequences for $b$ and $y$. This would also reveal other solutions such as $a=b=-1$ and $x$ and $y$ both odd integers, or $a=-1$ and $b=1$ and $x$ even, or $a=1$ and $b=-1$ and $y$ even.2012-12-14
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    @Henry: I know exactly what I’ve done. I never said that it was a complete solution: it very obviously isn’t, and it wasn’t intended to be. It was intended to show the OP how to deal with the main issue; having taught my share of precalculus courses, I’d bet that that was the point of the problem. I’d also bet that there’s a background assumption, possibly even stated explicitly at some point, that $a,b>0$.2012-12-14
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    Thanks a lot for your answers, Brian your approach is quite unique thank you! I checked the answer at the back of the textbook however and it hint at "logs to base a" being taken, what would be the purpose of this? That's what I was trying to get my head around.2012-12-14
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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