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How do you solve #logx+log(x+21)=2#?

1 Answer

Nov 23, 2016

#x=25#

Explanation:

#logx(x+21)=2#
#log(x^2+21x)=2#
#x^2+21x=10^2#
#x^2+21x-100=0#
#(x-4)(x+25)=0#
#x=4# and #x=-25#
#x=-25# is extraneous
#x=4#

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