Question 863604: How to solve 2log(x+1) - log (x+3) = log1
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! rules of log that apply (rule numbers are mine created for this problem only):
rule 1:
a*log(b) = log(b^a)
rule 2:
log(a) - log(b) = log(a/b)
rule 3:
if log(a) = log(b) then a = b
rule 4:
you cannot take the log of a negative number
original problem:
2log(x+1) - log(x+3) = log(1)
apply rule 1 to get:
log((x+1)^2) - log(x+3) = log(1)
apply rule 2 to get:
log((x+1)^2)/(x+3)) = log(1)
apply rule 3 to get:
(x+1)^2/(x+3) = 1
multiply both sides of this equation by (x+3) to get:
(x+1)^2 = (x+3)
subtract (x+3) from both sides of this equation to get:
(x+1)^2 - (x+3) = 0
simplify by performing indicated operations to get:
x^2 + 2x + 1 - x - 3 = 0
simplify by combining like terms to get:
x^2 + x - 2 = 0
factor to get:
(x+2) * (x-1) = 0
solve for x to get:
x = -2
x = 1
apply rule 4 to reject x = -2 as a valid answer because 2log(x+1) becomes 2log(-1).
your solution is x = 1.
substitute for x in the original equation to confirm this solution is good.
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