# precalculus

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log5 (x+3)= 1- log5 (X-1) where do I start?

• precalculus -

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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