Help with Maths question?

Question

If 2^x + 2^(1-x) - 3 = 0
How do I find the value of x?
If any one knows can they post the solution
I know it involves logs.

Thanks
The answer stated is x=1 or x=0 but im looking for the solution

I appreciate your reasoning but it is indeed a log problem as it states to use logarithms to solve

Answer 1

2^(1-x)=2/2^x
rewriting the equation,
2^x+2/2^x-3=0
2^2x+2-3*2^x=0
substitute 2^x=t
t^2-3t+2=0
(t-2)(t-1)=0
t=2 or 1
2^x=2,x=1
2^x=1, x=0
so x=0 or 1

Answer 2

I don't think it is really a log problem, it is a logic problem

x has to be one or less. if x was greater than one, then 2^(1-x) would be less than one and there won't be a solution. If x was 2, the equation would be 4 - something less than one - 3 = 0.
There is nothing less than 1 from which I can subtract 3 to get 0.

Looking at the numbers less than one, knowing that I can't use negative numbers, I see that the choices are 0 or 1. Substituting those for x, I see that 0 is the only thing that works.

Answer 3

2^x + 2^(1-x) - 3 = 0
2^x + 2^(1-x) = 3
now take log of each side of equation to get:
ln 2^x + ln 2^(1-x) = ln 3
x*ln2 +(1-x)*ln2 =ln 3
x+1-x = ln3/ln2
1= ln 3/ln 2 which is not true, so the traditional method of using logarithms for this problem does not seem to work.

Other than trial and error, I don't see how to solve this equation.