solve the equation 2^1-x=4^x?

2^(1 - x) = 4^x

Write it like this:

2^(1 - x) = (2²)^x
2^(1 - x) = 2^(2x)

Therefore we have that:

1 - x = 2x
3x = 1
x = 1/3

Hope this helps!

Question is incorrect. Should be :-
2^(1 - x) = 4^x
2^(1 - x) = 2^(2x)
1 - x = 2x
1 = 3x
x = 1 / 3

2^(1-x) = 4^x

Multiply both sides by 2^x

2^(1-x+x) + 2^1 = 4^x * 2^x = ((2^2)^x 2^x = (2^2x) * 2^x = 2^3x

Repeating with out the middle steps:

2^1 = 2^3x
so 3x = 1
x = 1/3 <<<<<<

i assume you mean 2^(1-x), not 2^1-x.

Take the log of both sides, so you get

(1-x) ln2 = x ln4

ln2 = x ln2 + x ln4 = x ln8

x = ln 2/ln 8

2^(1 - x) = 4^x
4^[(1 - x)/2] = 4^x
(1 - x)/2 = x
2x = 1 - x
2x + x = 1
3x = 1
x = 1/3

2^(1-x)=(2^2)^x=2^(2x)

1-x=2x -3x=-1 x=-1/-3=1/3