Comments on: HOW would I do this natural log problem?

By: Abhinav M

Abhinav M — Fri, 23 Nov 2007 02:45:59 +0000

ln(x+1)-lnx=1
=>ln[ (x+1)/x]=1
since lnm-lnp=ln(m/p)
=>(x+1)/x=e
=> x+1=ex
=>x=1/(e-1)
Now u can calculate by calculator
approx. comes out to be 1.39

By: Puzzling

Puzzling — Wed, 21 Nov 2007 19:41:31 +0000

Remember this rule of logs:
log(a) - log(b) = log(a/b)

ln((x+1) / x)) = 1

Raise both sides upon the base e:
(x + 1) / x = e

Multiply both sides by x:
x + 1 = ex

Subtract x from both sides:
1 = ex - x

Factor out an x:
1 = (e - 1)x

Divide both sides by (e-1)
x = 1 / (e-1)

By: hustolemyname

hustolemyname — Mon, 19 Nov 2007 01:52:51 +0000

yes usual log properties apply whatever the base

(x+1)/x = e
x = 1/(e-1)

By: Mich

Mich — Sun, 18 Nov 2007 20:12:27 +0000

you need to combine the two ln to one then rewite as an exponential equation use this rule log(a) - log(b) = log(a/b)
combining we get ln((x+1)/x) = 1
then rewite as an exponential
e^1 = (x+1)/x mul both sides by x to clear fractions
e^1(x) = x+1 put all x on one side
e(x) - x = 1 factor out the x
x(e - 1) = 1
x = 1/(e - 1)
hope this helped