5
$\begingroup$

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...

  • 1
    Where does this equation come from? – 2017-02-22
  • 2
    $x=0, x=2$ tis a solution. – 2017-02-22
  • 0
    have you tried plotting it – 2017-02-22
  • 1
    $0$ is obvious and WA got $2$ too. – 2017-02-22

2 Answers 2

8

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.

  • 2
    Love this use of FLT ! – 2017-02-22
  • 1
    Can we use FLT here when there is no guarantee of $x$ being an integer? – 2017-02-22
  • 4
    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. – 2017-02-22
5

Hint: your equation can be factorized as $$\left(2^x-3^x\right) \left(2^{2 x}+3^x-5^x\right)=0$$

  • 1
    How did you get this? I mean, it doesn't seem obvious. – 2017-02-22
  • 0
    I know this equation from Mathlinks – 2017-02-22
  • 1
    @S.C.B We can write the equation as: $$2^{3x} + 2^x3^x +3^x5^x = 2^{2x}3^x + 3^{2x} + 2^x5^x $$ $$\Rightarrow 2^x2^{2x} +2^x3^x-2^x5^x = 3^x2^{2x} +3^x3^x-3^x5^x $$ – 2017-02-22
  • 0
    i think this doesn't help at all – 2017-02-22
  • 0
    @Dr.SonnhardGraubner What is Mathlinks? – 2017-02-22
  • 0
    see here www.mathlinks.ro – 2017-02-22
  • 0
    $$x=0$$ or $$x=2$$ are the only solutions – 2017-02-22