Algebra Log problem (one more) solve 4 ^ x^3 = 2 ^x?

Question

the x is a superscript to the 4 and the 2 is a superscript to the x =
2 with a x superscript,
thanks if you can help me.

Answer 1

4^(x^3) = 2^x

2^(2x^3) = 2^x

2x^3 = x

2x^3 - x = 0

x(2x^2-1) = 0

x(sqrt2 x - 1)(sqrt2 x + 1) = 0

x = 0

x = (sqrt 2) / 2

x = -(sqrt 2) / 2

Answer 2

Solve for x.

4^ x^3 = 2 ^x

4^(x^3) = 2^x
(2²)^(x^3) = 2^x
2^(2x^3) = 2^x

Take the log to the base 2 of both sides

2x³ = x
2x³ - x = 0
x(2x² - 1) = 0
x(√2x + 1)(√2x - 1) = 0

x = 0, ±1/√2

Answer 3

oh yeah, forget everything but x^3 and x and hm.. I think you gotta hm..

first power up the x^3 so it's 2x^3 and then it's 2x^3=x and

then it's 2x^3-x=0 so then it'd be x(x^2-1) and that'd mean that hm.. x=1 I think. no wait, that's not right, x=1/2.

4^x^1/2=2^1/2 would mean that 4 hm... not sure

Answer 4

take the log of each side of the equation:
ln (4^x^3) = ln (2^x), which reduces to:
(x^3) (ln 4) = x (ln 2)

divide each side by x:
(x^2) (ln 4) = ln 2

since ln 4 = ln (2^2), ln 4 = 2 (ln 2). thus,

(x^2) (2) (ln 2) = ln 2
dividing each side by (ln 2), and then dividing each by 2:

x^2 = 1/2. taking the square root of each side,
x = square root of 1/2, or 1/square root of 2.