Posted by james on Monday, March 19, 2012 at 4:23am.
use your log rules:
log (AB) = logA + logB
log (A/B) = logA - logB
log3 [(x-5)(x+3)] = 2
by definition of logs
(x-5)(x+3) = 3^2 = 9
x^2 - 2x -24=0
(x-6)(x+4) = 0
x = 6 or x=-4 , but x=-4 would make log3(x+3) undefined, so
x = 6
log7(x) + log7(x-1) - log7(2x) = 0
log7[x(x-1)/(2x)] = 0
log7[(x-1)/2] = 0
(x-1)/2 = 7^0
(x-1)/2 = 1
x = 3
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