How do you solve for x in #3/2log_7x=log_7 125#?

1 Answer
Gió
Feb 11, 2017

I found #x=25#

Explanation:

Let us use a property of logs to get rid of the #3/2# as:
#log_7(x)^(3/2)=log_7(125)#
the logs are equal if the arguments are equal. So we need that:
#x^(3/2)=125#
let us take the exponent of #2/3# on both sides:
#x^(3/2*2/3)=125^(2/3)#
#x=125^(2/3)#
#x=root3(125^2)=25#

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