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 #