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

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Answer 1 · 2016-01-25T20:10:46

The answer is #x=25#

#logx+log4=2#

First we use the identity which says that: #loga+logb=log(a*b)#

#log(4x)=2#

Then we change the number on the right side to a logarythm using a formula:

#a=log_b(b^a)#

#log(4x)=log100#

Now we can skip the logarythm signs because both sides are single logarythms with the same base.

#4x=100#

#x=25#