Asked 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)
ln(x-5/x^2 - 1) - ln (x^2 - 2x -15/x + 1)
Answers
Answered by
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)
= 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)
Answered by
James
Thank you!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.