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I have tried solving the following equation by using exponential properties and logarithms, but can not find some link between all of the terms:

$$6\times3^{2x}-13 \times6^x +6\times 2^{2x}=0$$

EDIT: After some research it resulted that the equation was wrongly printed, and for that I am truly sorry to all of you who spent your time trying to figure it out. The exact form was the one as described in the comments.

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    $\begingroup$ This has an approximate numerical solution for $x=0.813332$. Are you sure that the question is correct? $\endgroup$
    – John McGee
    Jan 16, 2015 at 21:28
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    $\begingroup$ Do you maybe mean $13\cdot 6^x$? $\endgroup$
    – ryagami
    Jan 16, 2015 at 21:37
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    $\begingroup$ @ryagami: For your speculated equation, the answer is $(3\cdot2^x-2\cdot3^x)(2^{x+1}-3^{x+1})=0 \implies x=\pm1$. $\endgroup$ Jan 16, 2015 at 21:40
  • $\begingroup$ @barakmanos Yes, when it's what I wrote, it factors rather nicely. $\endgroup$
    – ryagami
    Jan 16, 2015 at 21:41
  • $\begingroup$ @ryagami: It's actually a nice riddle by itself: "place an operation somewhere within the equation so that the solutions are $\pm1$". $\endgroup$ Jan 16, 2015 at 21:44

1 Answer 1

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From your equation you get: $(3^{2x}+2^{3x})/{136^x}=1/6\Rightarrow (9/136)^x+(4/136)^x=1/6$.

The second member is constant while the first one is a function in $x$: $f(x)=(3^{2x}+2^{3x})/{136^x}$.

The function is strictly decreasing, so the equation has at most one solution. Checking with Wolphram Alpha you get a numerical solution. You can actually state that for $x\geq 0$ it has exactly one positive solution, since $f(0)=2$ and $\lim_{x\rightarrow\infty} f(x)=0$. (if you have studied a bit of analysis).

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