How do you solve #log_5 10 * log_5 x = log_5 100#?

1 Answer
Gió
Jul 3, 2016

I found: #x=25#

Explanation:

I would try to write it in a different way as:
#log_5(10)log_5(x)=log_5(10)^2#
using a property of logs (expoent of the argument):
#log_5(10)log_5(x)=2log_5(10)#
rearrange:
#log_5(x)=(2cancel(log_5(10)))/cancel(log_5(10))#
so:
#log_5(x)=2#
using the definition of log:
#x=5^2=25#

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