# pre calculus

posted by James

Write the expression below as a single logarithm. Need to know how to work this!
ln(x-5/x^2 - 1) - ln (x^2 - 2x -15/x + 1)

1. Reiny

ln(x-5/x^2 - 1) - ln (x^2 - 2x -15/x + 1)
= ln(x-5) - ln(x^2 - 1) - (ln(x-5)(x+3) - ln(x+1) )
= ln(x-5) -( (ln (x+1) + ln(x-1) ) - ( ln(x-5) + ln(x+3) - ln(x+1) )
= ln(x-5) - ln(x+1) - ln(x-1) - ln(x-5) - ln(x+3) + ln(x+1)
= -( ln(x-1) + ln(x+3) )
= -ln( (x-1)(x+3) )
or -ln(x^2 + 2x - 3)

2. James

Thank you!

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