Question 372076
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Hi,
log base3 (x-8)+ log base 3 x = 2

*[tex \large\ \ log_bx + log_by = log_b(xy) ]
*[tex \large\ \ log_3(x-8) + log_3x = log_3((x-8)x) ] = 2
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]
3^2 = (x-8)x
solving for x
9 = x^2 - 8x
0 = x^2 - 8x - 9
factoring
(x-9)(x+1) = 0
(x-9)= 0  x = 9
(x+1)= 0  x = -1 Cannot use 
x = 9