How do you evaluate #log_a x = 3/2log_a 9 + log_a 2#?
3 Answers
We will use the following log properties:
Explanation:
Practice exercises

Solve for x in the following equation. Round your answers to the nearest hundredth:
#log_n2x = 1/3log_n64 2log3# 
A commonly used logarithm rule is if
#a = b > loga = logb# . Solve for x in#2^(2x  3) = 3^(3x)#
Good luck!
x = 54
Explanation:
using the
#color(blue)" laws of logarithms "#
#• logx + logy = logxy #
#• logx  logy = log(x/y) #
#• logx^n = nlogx hArr nlogx = logx^n #
#• "If " log_b x = log_b y rArr x = y #
#rArr log_a x = log_a 9^(3/2) + log_a 2 # [ now
#9^(3/2) = (sqrt9)^3 = 3^3 = 27 # ]
#rArr log_a x = log_a (27xx2) = log_a 54 # hence x = 54
Explanation:
We are given
Anyways, I now have my equation as
This now gives us
Applying that logic, I take