If I have 2^(3-x) + 2^x = 9 how do I isolate the x's?

Question

anonymous · Answer

Bring down logs and use logx^n = nlogx

haseeb96 · Answer

2^3.2^-x + 2^x =9
$$\frac{ 2^3 }{ 2^x }+2^x = 9$$
$$\frac{ 8+4x^2 }{ 2^x }=9$$
8+4x^2=18^x 
4x^2-18^x+8=0
taking two common  from this equation
2(2x^2-9x+4)=0
2x^2-9x+4=0
using factorized method 
2x^2-1x-8x+4=0
x(2x-1)-4(2x-1)=0
(x-4)(2x-1)=0
either
x-4=0              or                 2x-1=0
x=4          or                  2x=1
                                        x=1\2
this is the correct answer

anonymous · Answer

Thanks for the help I actually think the answer is 0 and 3 though

haseeb96 · Answer

no, it is wrong

anonymous · Answer

no its right, 9 * 2^x is not 18^x otherwise your math was right. I saw what I did wrong and it is 0 and 3. You can check it if you plug it back into the equation.