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)
= 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)
Thank you!
To write the given expression as a single logarithm, we can use the logarithmic identity:
ln(a) - ln(b) = ln(a/b)
Let's apply this identity to the given expression:
ln((x - 5)/(x^2 - 1)) - ln((x^2 - 2x - 15)/(x + 1))
Applying the logarithmic identity, we can rewrite this as:
ln(((x - 5)/(x^2 - 1))/((x^2 - 2x - 15)/(x + 1)))
Now, let's simplify the expression inside the logarithm:
((x - 5)/(x^2 - 1))/((x^2 - 2x - 15)/(x + 1))
To simplify this further, we can rewrite the division of fractions as multiplication, by taking the reciprocal of the second fraction:
((x - 5)/(x^2 - 1)) * ((x + 1)/(x^2 - 2x - 15))
Now, we can cancel out the common factors in the numerator and denominator:
((x - 5)/(x + 1)) * ((x + 1)/(x - 5))
Notice that (x + 1)/(x + 1) and (x - 5)/(x - 5) both equal 1. Therefore, this simplifies to:
(x - 5)/(x + 1)
Now, we can rewrite the expression as a single logarithm:
ln((x - 5)/(x + 1))
And there you have it! The given expression, ln((x-5)/(x^2 - 1)) - ln (x^2 - 2x -15/x + 1), can be written as a single logarithm: ln((x - 5)/(x + 1)).