How do you solve 8^sqrtx = 4^root3 x8x=43x?

1 Answer
Feb 23, 2017

{0, 64/729}{0,64729}

Explanation:

Rewrite in base 22.

(2^3)^sqrt(x) = (2^2)^root(3)(x)(23)x=(22)3x

2^(3sqrt(x)) = 2^(2root(3)(x)23x=223x

We don't need the bases anymore.

3sqrt(x) = 2root(3)(x)3x=23x

3x^(1/2) = 2x^(1/3)3x12=2x13

Put both sides to the 6th power to get rid of fractional exponents (we take the 6th power because the LCM of 22 and 33 is 66).

(3x^(1/2))^6 = (2x^(1/3))^6(3x12)6=(2x13)6

729x^3 = 64x^2729x3=64x2

729x^3 - 64x^2 = 0729x364x2=0

x^2(729x - 64) = 0x2(729x64)=0

x = 0 and 64/729x=0and64729

Hopefully this helps!