How do you solve #log_3x-2log_3 2=3log_3 3 #?

1 Answer
Gió
Feb 29, 2016

I found : #x=108#

Explanation:

We can use the properties of logs that tell us:
#logx-logy=log(x/y)#
and also:
#logx^a=alogx#
in our case we get:
#log_3x-log_3(2^2)=log_3(3^3)#
then:
#log_3(x/2^2)=log_3(27)#
if the logs must be equal the arguments must be as well. So, we can write:
#x/4=27#
#x=4*27=108#

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