How do you solve #log _4 (x)+ log _4 (x^2) =5#?
1 Answer
Dec 31, 2015
Use exponentiation with
#x = 4^(5/3) ~~ 10.0793684#
Explanation:
#x^3 = 4^(log_4(x^3)) = 4^(log_4(x * x^2)) = 4^(log_4(x) + log_4(x^2)) = 4^5#
So:
#x = root(3)(4^5) = (4^5)^(1/3) = 4^(5/3)#
