precalculus
posted by Sara .
log5 (x+3)= 1 log5 (X1) where do I start?

start by collecting logs:
log5 (x+3) + log5(x1) = 1
now recall that sum of logs is log of product
log5 [(x+3)(x1)] = 1
raise 5 to the power of both sides:
(x+3)(x1) = 5
x^2 + 2x  3 = 5
x^2 + 2x  8 = 0
(x+4)(x2) = 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 (21)
log5 5 = 1  log5 1
1 = 1  0