How do you solve #-2=log_x (1/100)#?

1 Answer
Dec 21, 2015

#x= 10 #

Explanation:

#-2 = log_x (1/100)#

There are many ways you can solve this problem. The best approach is to convert the given equation to exponent form.

Rule : #log_b (a) = k => a= k^b#

Using the rule we can get

#x^-2 = 1/100#
#x^-2 =1/10^2#
#x^-2 = 10^-2# Rule #1/a^m = a^-m #

#x= 10 # The final answer.

Alternate approach

#-2 = log_x (1/100)#
#-2 = log_x (10^-2)#
#-2 = -2 log_x (10) # Rule #log(A)^n = n log(A)#
# 1 = log_x (10) #
#x =10 # because of the rule #log_a (a) = 1# x has to be 10.