How do you solve #log 4x = 2#?

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Answer 1 · 2018-06-26T14:34:14

Rewrite in in exponential form!

Note that logarithms with a base of 10 (#log_10 x#) are commonly rewritten without it (#logx#).

#log_a b=x# is the same as #a^x=b#. Let's apply this to our equation:

#log_10 4x=2#
#10^2=4x#
#100=4x#
#25=x#

Hope this helped!

Answer 2 · 2018-06-26T14:36:11

x = 25

#log(4x)= 2#
Since the base is not mentioned we assume it to be 10.
#4x=10^2#
#x=100/4=25#