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How do you solve #log_6(x + 19) + log_6(x) = 3#?

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Konstantinos Michailidis

Oct 4, 2015

We must have #x>0# hence

#log_6(x + 19) + log_6(x) = 3=>log_6 x(x+19)=3=> x(x+19)=6^3=>x^2+19x-216=0=>x^2+27x-8x-216=0=> (x-8)*(x+27)=0=>x=8 or x=-27#

But #x=-27# is rejected so #x=8# is the only solution

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