Solve exponential equation $6\times3^{2x}-13\times 6^x +6\times 2^{2x}=0$

Question

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.

This has an approximate numerical solution for $x=0.813332$. Are you sure that the question is correct? — John McGee, Jan 16 '15 at 21:28
@ryagami: For your speculated equation, the answer is $(3\cdot2^x-2\cdot3^x)(2^{x+1}-3^{x+1})=0 \implies x=\pm1$. — barak manos, Jan 16 '15 at 21:40
@barakmanos Yes, when it's what I wrote, it factors rather nicely. — ryagami, Jan 16 '15 at 21:41
@ryagami: It's actually a nice riddle by itself: "place an operation somewhere within the equation so that the solutions are $\pm1$". — barak manos, Jan 16 '15 at 21:44

score 4 · Answer 1 · answered Jan 18 '15 at 00:51

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).

Solve exponential equation $6\times3^{2x}-13\times 6^x +6\times 2^{2x}=0$

1 Answers1