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 Subject: Solve for x Category: Reference, Education and News > Homework Help Asked by: xkyroutx-ga List Price: \$4.00 Posted: 10 Dec 2005 19:14 PST Expires: 09 Jan 2006 19:14 PST Question ID: 604248
 ```ln(x+1)-lnx=e I have to solve for x. The problem I'm having is the - throws me off. This is what I've come up with. I distribute the lnto the (x+1) to come up with lnx+ln1-lnx=e the lnx's cancel eachother out and therefore I cannot solve for x. I'm looking for some instructions as to how to solve this.```
 ```Hi xkyroutx-ga, Natural logarithm (ln) can not be distributed like what you have done: you can not separate ln(x+1) into ln(x)+ln(1). The following are some logarithmic identities that you should know: 1. log(A*B) = log(A)+log(B) 2. log(A/B) = log(A)-log(B) 3. log(A^B) = B * log(A) Now, let's look at your equation: ln(x+1)-ln(x) = e By identity #2 above, you can simplify ln(x+1)-ln(x) into ln((x+1)/x), giving: ln((x+1)/x) = e The next is to take the exponent of e on both sides: e^(ln((x+1/x) = e^e (x+1)/x = e^e, x not equal to 0 now we multiply both sides by x, since x is not 0: x+1 = x * e^e and subtract x from both sides: 1 = x * e^e - x 1 = x(e^e - 1) and divide both sides by e^e - 1: x = 1/(e^e - 1) Now to check the answer: x + 1 = 1/(e^e-1) + 1 x = 1/(e^e-1) (x+1)/x = (1/(e^e-1) + 1)*(e^e-1) (x+1)/x = 1 + e^e - 1 = e^e ln ((x+1)/x) = ln (e^e) = e So your answer is x = 1/(e^e - 1). Reference: Logarithmic identities: http://en.wikipedia.org/wiki/Logarithm#Easier_computations I hope that this was clear. If you need clarification, please use the request for clarification feature before rating this answer. secret901-ga```