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What are the solutions of this equation? Or at least in which interval are they? $$6^x+8^x+15^x=9^x+10^x+12^x$$ I tried to find an increasing function, or use some inequalities but I got nothing out of it...

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    $\begingroup$ Where does this equation come from? $\endgroup$ – Henning Makholm Feb 22 '17 at 11:15
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    $\begingroup$ $x=0, x=2$ tis a solution. $\endgroup$ – S.C.B. Feb 22 '17 at 11:17
  • $\begingroup$ have you tried plotting it $\endgroup$ – mercio Feb 22 '17 at 11:17
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    $\begingroup$ $0$ is obvious and WA got $2$ too. $\endgroup$ – Zubzub Feb 22 '17 at 11:17
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Let $a=3^x$ and $b=2^x$ and $c=5^x$.

Then we have that $$ab+b^3+ac=a^2+bc+ab^2$$ $$ab+b^3+ac-a^2-bc-ab^2=0$$ $$a(b-a)+b^2(b-a)-c(b-a)=0$$ $$(b-a)(a+b^2-c)=0$$

Now $a=b \implies x=0$ and $a+b^2-c=0 \implies 3^x+4^x=5^x$ which gives solution for $x=2$ only and not for any higher integer $x$ by Fermat's Last Theorem.

So these are the $2$ solutions.

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    $\begingroup$ Love this use of FLT ! $\endgroup$ – Zubzub Feb 22 '17 at 11:26
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    $\begingroup$ Can we use FLT here when there is no guarantee of $x$ being an integer? $\endgroup$ – didgogns Feb 22 '17 at 11:35
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    $\begingroup$ No, but there's a simple way around this: Dividing both sides by $5^x$ gives $\left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x = 1$, but the quantity on the l.h.s. is a strictly decreasing function of $x$, so the equation has at most one solution. $\endgroup$ – Travis Feb 22 '17 at 11:59
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Hint: your equation can be factorized as $$\left(2^x-3^x\right) \left(2^{2 x}+3^x-5^x\right)=0$$

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    $\begingroup$ How did you get this? I mean, it doesn't seem obvious. $\endgroup$ – S.C.B. Feb 22 '17 at 11:19
  • $\begingroup$ I know this equation from Mathlinks $\endgroup$ – Dr. Sonnhard Graubner Feb 22 '17 at 11:20
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    $\begingroup$ @S.C.B We can write the equation as: $$2^{3x} + 2^x3^x +3^x5^x = 2^{2x}3^x + 3^{2x} + 2^x5^x $$ $$\Rightarrow 2^x2^{2x} +2^x3^x-2^x5^x = 3^x2^{2x} +3^x3^x-3^x5^x $$ $\endgroup$ – Rohan Feb 22 '17 at 11:21
  • $\begingroup$ i think this doesn't help at all $\endgroup$ – Dr. Sonnhard Graubner Feb 22 '17 at 11:22
  • $\begingroup$ @Dr.SonnhardGraubner What is Mathlinks? $\endgroup$ – S.C.B. Feb 22 '17 at 11:22

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