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Let $a, b \in (0, \infty)$, $a<b$. Prove that the equation $$\left(\frac{a+b}{2}\right)^{x+y}=a^xb^y$$ has at least one solution in $(a, b)$. Some suggestions? Thanks.

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Do you want to prove that for any $x,y \in \mathbb{R}$, there exists $a,b \in (0,\infty)$ such that $((a+b)/2)^{x+y} = a^x b^y$? –  user17762 Dec 3 '12 at 21:06
    
@Marvis: not really. The text of the problem is somewhat different. For instance, I think it's possible you may choose $x, y \in R$ such that there is no a,b so that $((a+b)/2)^{x+y} = a^x b^y$. (I just suppose that, but I might be wrong). The text above is the original text of the problem the way I received it. –  Chris's sis Dec 3 '12 at 21:16
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The solutions are pairs $(x,y) \in \mathbb{R}^2$, yet the solution space is supposed to be an interval in $\mathbb{R}$. I'm not sure what is meant. It is probably best to get clarification from the original source as to what the problem actually is asking you to do. –  Michael Joyce Dec 3 '12 at 21:24
    
Is $(a,b)$ an open interval or a pair? –  glebovg Dec 3 '12 at 21:32
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Well you definitely want to start by taking the logarithm of both sides ... –  Michael Joyce Dec 3 '12 at 21:37
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2 Answers 2

up vote 6 down vote accepted

This is a direct usage of the intermediate value theorem.

Let $f(t)=a^{1-t}b^t$. Then $f(t)$ is continuous and $f(0)=a$ and $f(1)=b$. Since $\frac{a+b}2$ is between $a$ and $b$, there must be a $t_0\in[0,1]$ such that $$f(t_0)=\frac{a+b}2$$ Let $x=1-t$ and $y=t$.

Indeed, we could just solve for $t_0$ by writing $f(t)=a\left(\frac{b}a\right)^t$ and yielding $t_0=\log_{b/a} \frac{a+b}{2a}$

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Too awesome! Thanks! –  Chris's sis Dec 3 '12 at 22:10
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The equation can be solved for $x/y$ in terms of $b/a$ as $$ \frac{x}{y} = \frac{\ln(b/a)}{\ln\left(\dfrac{1+b/a}{2}\right)} - 1 $$

The statement that there exist $x,y \in (a,b)$ that satisfy the equation is equivalent to the assertion that for all $t > 1$,

$$ 1/t \le \frac{\ln(t)}{\ln\left(\dfrac{1+t}{2}\right)} - 1 \le t $$

Let $$\eqalign{f(t) &= t \ln \left(\dfrac{1+t}{2}\right)\cr g(t) &= \ln(t) - \ln \left(\dfrac{1+t}{2}\right)\cr h(t) &= \dfrac{1}{t} \ln \left(\dfrac{1+t}{2}\right)\cr}$$ Then we want to show that $f(t) > g(t) > h(t)$ for $t > 1$. As $t \to 0$, all three have limit $0$. So it suffices to show that $f'(t) - g'(t) > 0$ and $g'(t) > h'(t)$ for $t > 1$.

In fact we have $$f'(t) - g'(t) = \ln\left(\dfrac{1+t}{2}\right) + 1 - \frac{1}{t} \ge 1 - \frac{1}{t} > 0$$ $$ g'(t) - h'(t) = \frac{1}{t^2} \ln\left(\dfrac{1+t}{2}\right) > 0$$

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thank your for your solution (+1) –  Chris's sis Dec 3 '12 at 22:10
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