SOLUTION: Ln(x-1)+Ln(x+3)=Ln5
Cannot figure out how to solve, please help
Algebra.Com
Question 201009: Ln(x-1)+Ln(x+3)=Ln5
Cannot figure out how to solve, please help
Found 2 solutions by solver91311, RAY100:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Given the same base, the sum of the logs is the log of the product, and if two logs to the same base are equal, then their arguments must be equal.
Symbolically,
 + \log_b(y) = \log_b(xy))
and
 = \log_b(y) \ \ \Leftrightarrow\ \ x = y)
So:
 + \ln(x + 3) = \ln\left((x-1)(x+3)\right) = \ln(x^2 + 2x - 3))
and therefore
= \ln(5))
so
Solve the quadratic by ordinary methods -- this one will factor neatly.
John
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
ln(x-1) + ln (x+3) = ln 5
.
remember lna+lnb =ln(ab)
.
ln{(x-1)(x+3)} = ln5
.
(x-1)(x+3) =5
.
x^2 +2x -3 =5
.
x^2 +2x -8 =0
.
(x+4)(x-2) = 0
.
x = -4, x=2
.
check
.
(x=2),,,,ln { (2)-1} +ln{(2) +3} = ln5,,,,ln 1 +ln 5 =ln5,,,,,ok
.
(x=-4),,ln{ (-4)-1} +ln{(-4)+3} =ln5,,ln-5 +ln-1 =ln5,,,ln(-5*-1) =ln5,,,ok
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