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How do you solve #log x^2 + log 25 = 2#?

1 Answer

Konstantinos Michailidis

Sep 11, 2015

The answer is #x=2#

Explanation:

We have that

#logx^2+log25=2=>log25x^2=log100=>25x^2=100=>x^2=4=>x=2,x=-2#

But #x>0# hence #x=2#

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