Question 375445
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Hi,
Solving
{{{ ln(x-6)+ln(x+3)=ln22 }}}
{{{ ln(x-6)+ln(x+3)-ln22=0 }}}
{{{ ln((x-6)(x+3)/22)=0 }}}
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
10^0 = [(x-6)(x+3)]/22
1 = [(x-6)(x+3)]/22
22 = (x-6)(x+3)
22 = x^2 -3x - 18
0 = x^2 -3x - 40
factoring
(x-8)(x+5)= 0
(x+5) = 0
x = -5  extraneous solution
(x-8)=0
x = 8