How do you solve #log x=3/2 log 9+log 2#?

1 Answer
Mar 3, 2016

x = 54

Explanation:

using the following#color(blue)" laws of logarithms "#

• logx + logy = logxy

• log#x^n hArr nlogx #

and if #log_bx = log_by rArr x = y #

#rArr logx = log9^(3/2) + log2

[now #[9^(3/2) = (sqrt9)^3 = 3^3 = 27 ]#

#rArr logx = log(27xx2) = log54 rArr x = 54 #