Thread: How to solve this powers problem

1. How to solve this powers problem

This equation was posted up by OnlyMath but got deleted because he posted 3 problems in 1 thread. He has not reposted it so I am because I'm curious as to how to solve it.

$\displaystyle{2(4^x) + 6^x = 9^x}$ Find x.

I already know the answer is:

$\displaystyle{x = \dfrac{log2}{log(\frac{3}{2})}}$

$\displaystyle{x \approx{1.70951}$

I have tried really hard to try and solve it, but I get stuck because there is a plus between the powers.
I can't go $\sqrt[x]{-6^x}+\sqrt[x]{9^x}$ because it doesn't work. You have to go $\sqrt[x]{-6^x + 9^x}$ which doesn't help in solving x.

Can someone show me a step in the right direction?

2. Originally Posted by Educated
This equation was posted up by OnlyMath but got deleted because he posted 3 problems in 1 thread. He has not reposted it so I am because I'm curious as to how to solve it.

$\displaystyle{2(4^x) + 6^x = 9^x}$ Find x.

I already know the answer is:

$\displaystyle{x = \dfrac{log2}{log(\frac{3}{2})}}$

$\displaystyle{x \approx{1.70951}$

I have tried really hard to try and solve it, but I get stuck because there is a plus between the powers.
I can't go $\sqrt[x]{-6^x}+\sqrt[x]{9^x}$ because it doesn't work. You have to go $\sqrt[x]{-6^x + 9^x}$ which doesn't help in solving x.

Can someone show me a step in the right direction?
Arrange the equation to get

$
(3^x)^2-3^x\cdot 2^x-2(2^x)^2=0$

$
(3^x-2\cdot 2^x)(3^x+2^x)=0$

The second factor is undefined since $a^x<0.$

As for the first factor,

$3^x=2^{x+1}$

$\frac{x}{x+1}=\frac{\log 2}{\log 3}$

$x=\frac{\log 2}{\log \frac{3}{2}}$

3. Oh I got it, I would have never thought of trying to factorise it.

But the final step seems confusing.
Instead I did this for the final step:

$3^x=2^{x+1}$

$(x)log{3}=(x+1)log{2}$

$(x)log{3}=(x)log{2} + (1)log{2}$

$(x)log{3}-(x)log{2}= log{2}$

$x(log{3}-log{2})= log{2}$

$x= \dfrac{log{2}}{(log{3}-log{2})}$

$x= \dfrac{log{2}}{log{(\frac{3}{2}})}$

Thanks!