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July 23, 2014

July 23, 2014

Posted by **Sara** on Monday, January 9, 2012 at 11:07am.

- precalculus -
**Steve**, Monday, January 9, 2012 at 11:32amstart 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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