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How do you solve for x in #log_3(x^(2)-2x)=1#?

1 Answer

Dec 31, 2015

#x_1=-1# and #x_2=3#

Explanation:

#log_3 (x^2-2x)=1#
#x^2-2x=3^1#
#x^2-2x-3=0#
#Delta = 4 +12 =16#
#x=(-b+-sqrt(Delta))/2=(2+-4)/2# => #x_1=-1# and #x_2=3#

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