Posted by strawberryfields on .
okay, this i cannot figure out anywhere.
2[ln(x)ln(x+1)ln(x1)]
evaluate into logarithm of a single quantity.

math 
drwls,
ln[x/(x^21)^2]

math 
strawberryfields,
would you mind telling me how you got it?
i don't get this problem..
thanks forthe answer though! 
math 
drwls,
Actually I should have written it as
ln{[x/(x^21]^2}
The ln(x+1)ln(x1) in the denominator is ln {1/[(x+1)(x1)] = ln [1/(x^21)]
Adding ln x puts the x in the numerator of [x/(x^21)]
The two in front of
2[ln(x)ln(x+1)ln(x1)]
is the same as taking the log of the square of [ln(x)ln(x+1)ln(x1)]