How do you solve 8^sqrtx = 4^root3 x8√x=43√x?
1 Answer
Feb 23, 2017
Explanation:
Rewrite in base
(2^3)^sqrt(x) = (2^2)^root(3)(x)(23)√x=(22)3√x
2^(3sqrt(x)) = 2^(2root(3)(x)23√x=223√x
We don't need the bases anymore.
3sqrt(x) = 2root(3)(x)3√x=23√x
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
(3x^(1/2))^6 = (2x^(1/3))^6(3x12)6=(2x13)6
729x^3 = 64x^2729x3=64x2
729x^3 - 64x^2 = 0729x3−64x2=0
x^2(729x - 64) = 0x2(729x−64)=0
x = 0 and 64/729x=0and64729
Hopefully this helps!