SOLUTION: I really need help solving this equation. PLEASE help? ln(x+1)-ln(x-2)=ln(5) 1. Solve equation by setting up algebraically. 2. Use appropriate properties of Logarithm to red

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I really need help solving this equation. PLEASE help? ln(x+1)-ln(x-2)=ln(5) 1. Solve equation by setting up algebraically. 2. Use appropriate properties of Logarithm to red      Log On


   

Question 410617: I really need help solving this equation. PLEASE help?
ln(x+1)-ln(x-2)=ln(5)
1. Solve equation by setting up algebraically.
2. Use appropriate properties of Logarithm to reduce the number of logarithmic terms to only one.
3. Solve logarithmic equation.
Answer by CharlesG2(834) About Me  (Show Source):
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"I really need help solving this equation. PLEASE help?
ln(x+1)-ln(x-2)=ln(5)
1. Solve equation by setting up algebraically.
2. Use appropriate properties of Logarithm to reduce the number of logarithmic terms to only one.
3. Solve logarithmic equation. "

ln(x + 1) - ln(x - 2) = ln(5)
logarithmic rule: ln(m) - ln(n) = ln(m/n)
ln((x + 1)/(x - 2)) = ln(5)
(x + 1)/(x - 2) = 5 (what was in parentheses on above line must equal)
x + 1 = 5(x - 2)
x + 1 = 5x - 10
-4x = -11
x = 11/4
check:
ln(11/4 + 1) - ln(11/4 - 2) = ln(5)
ln(11/4 + 4/4) - ln(11/4 - 8/4) = ln(5)
ln(15/4) - ln(3/4) = ln(5)
ln(15/4 * 4/3) = ln(5)
ln(15/3) = ln(5)
ln(5) = ln(5) , yes