How do you solve # ln x = 2 ln (x - 6)#?

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Answer 1 · 2016-07-10T19:04:43

#lnx - 2ln(x - 6) = 0#

#lnx - ln(x - 6)^2 = 0#

#lnx - ln(x^2 - 12x + 36) = 0#

#ln(x/(x^2 - 12x + 36)) = 0#

#x/(x^2 - 12x + 36) = e^0#

#x = 1(x^2 - 12x + 36)#

#0 = x^2 - 13x + 36#

#0 = (x - 9)(x - 4)#

#x = 9 and 4#

Checking the solutions in the original equation, we find that only #x = 9# works. Hence, the solution set is #{9}#.

Hopefully this helps!