write the expression as a sinle logarithm. express powers as factors.
In (x/x-1)+In(x+1/x)-In(x^2-1)
Please show work.
You MUST put brackets for the denominators to avoid ambiguity
I am sure you mean
ln ( x/(x+1) ) + ln( (x+1)/x ) - ln (x^2 - 1)
then
= lnx - ln(x+1) + (ln(x+1) - lnx - ln( (x+1)(x-1)
= lnx - ln(x+1) + ln(x+1) + lnx - ln(x+1) - ln(x-1)
= - 2ln(x+1)
or
ln [ (x/(x-1) ((x+1)/x) / ((x+1)x-1))
= ln (1/(x-1)^2)
= ln1 - ln(x-1)^2
= 0 - 2ln(x-1)
= -2ln(x-1)
To rewrite the expression as a single logarithm, we need to use the properties of logarithms, including addition, subtraction, and multiplication.
Starting with the given expression:
In(x/x-1) + In(x+1/x) - In(x^2-1)
First, we can use the quotient property of logarithms, which states that ln(a/b) is equal to ln(a) - ln(b).
Therefore, we can rewrite the expression as:
ln((x/(x-1)) * (x+1/x)) - ln(x^2-1)
Next, let's simplify the numerator and denominator separately.
For the numerator, we can simplify it by finding a common denominator:
x(x+1) / [(x-1)x]
This can be further simplified to:
(x(x+1))/(x(x-1))
Now, we can rewrite the expression as:
ln((x(x+1))/(x(x-1))) - ln(x^2-1)
Next, we can use the product property of logarithms, which states that ln(ab) is equal to ln(a) + ln(b).
Using the product property, we can rewrite the expression as:
ln(x(x+1))/(x(x-1)) - ln(x^2-1)
Finally, we can combine the two logarithms into a single logarithm using the subtraction property, which states that ln(a) - ln(b) is equal to ln(a/b).
Thus, the final expression, written as a single logarithm, is:
ln[(x(x+1))/(x(x-1)*(x^2-1))]