How do you solve #log x^2+1=5#?

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How do you solve #log x^2+1=5#?

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Answer 1 · 2016-03-21T19:21:07

#x=100#

Explanation:

#log x^2 +1 =5#
#logx^2=4#-> subtract 1 from both sides
#2logx=4#->Use property #log_bx^n=n*log_b x#
#log x=2#-> divide both sides by 2
#10^2=x#->use property #y=log_bx iff b^y=x#
#x=100#

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How do you solve #log x^2+1=5#?

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