okay, this i cannot figure out anywhere.
2[ln(x)-ln(x+1)-ln(x-1)]
evaluate into logarithm of a single quantity.
ln[x/(x^2-1)^2]
would you mind telling me how you got it?
i don't get this problem..
thanks forthe answer though!
Actually I should have written it as
ln{[x/(x^2-1]^2}
The -ln(x+1)-ln(x-1) in the denominator is ln {1/[(x+1)(x-1)] = ln [1/(x^2-1)]
Adding ln x puts the x in the numerator of [x/(x^2-1)]
The two in front of
2[ln(x)-ln(x+1)-ln(x-1)]
is the same as taking the log of the square of [ln(x)-ln(x+1)-ln(x-1)]
To evaluate the expression 2[ln(x) - ln(x+1) - ln(x-1)] into a single logarithm, let's simplify each term step by step:
1. Start with the expression: 2[ln(x) - ln(x+1) - ln(x-1)]
2. Using the properties of logarithms, we can rewrite the expression as the sum and difference of logarithms:
2[ln(x) - ln(x+1) - ln(x-1)] = 2[ln(x) - (ln(x+1) + ln(x-1))]
3. Now, let's use the product rule of logarithms, which states that ln(a) + ln(b) = ln(ab):
2[ln(x) - (ln(x+1) + ln(x-1))] = 2[ln(x) - ln((x+1)(x-1))]
4. Simplify the inside of the logarithm:
2[ln(x) - ln((x+1)(x-1))] = 2[ln(x) - ln(x^2 - 1)]
5. Further simplify by using the quotient rule of logarithms, which states that ln(a) - ln(b) = ln(a/b):
2[ln(x) - ln(x^2 - 1)] = 2[ln(x/(x^2 - 1))]
Therefore, the expression 2[ln(x) - ln(x+1) - ln(x-1)] is equivalent to 2[ln(x/(x^2 - 1))].
Note: Simplifying the expression to the form of a single logarithm may not always be possible, but in this case, it can be done as shown above.