Posted by Sara on Monday, January 9, 2012 at 11:07am.
start by collecting logs:
log5 (x+3) + log5(x-1) = 1
now recall that sum of logs is log of product
log5 [(x+3)(x-1)] = 1
raise 5 to the power of both sides:
(x+3)(x-1) = 5
x^2 + 2x - 3 = 5
x^2 + 2x - 8 = 0
(x+4)(x-2) = 0
x = 2 or -4
Since logs of negative numbers do not exist, we have to throw out x = -4 since it does not fit the original equation.
x=2 is our only solution.
check:
log5 (2+3) = 1 - log5 (2-1)
log5 5 = 1 - log5 1
1 = 1 - 0
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