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

Question

894 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-23T03:49:01

#{0, 64/729}#

Rewrite in base #2#.

#(2^3)^sqrt(x) = (2^2)^root(3)(x)#

#2^(3sqrt(x)) = 2^(2root(3)(x)#

We don't need the bases anymore.

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

#3x^(1/2) = 2x^(1/3)#

Put both sides to the 6th power to get rid of fractional exponents (we take the 6th power because the LCM of #2# and #3# is #6#).

#(3x^(1/2))^6 = (2x^(1/3))^6#

#729x^3 = 64x^2#

#729x^3 - 64x^2 = 0#

#x^2(729x - 64) = 0#

#x = 0 and 64/729#

Hopefully this helps!