Asked by Exponential
log (base 9) (x-2) + log (base 9) (x-2)=1
Solve algebraically.
Solve algebraically.
Answers
Answered by
Damon
2 log9 (x-2) = 1
log9 (x-2)^2 = 1
(x-2)^2 = 9^1 = 9
x^2 - 4x + 4 = 9
x^2 -4 x -5 = 0
(x-5)(x+1) = 0
x = 5 or x = -1
log9 (x-2)^2 = 1
(x-2)^2 = 9^1 = 9
x^2 - 4x + 4 = 9
x^2 -4 x -5 = 0
(x-5)(x+1) = 0
x = 5 or x = -1
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