HOW would I do this natural log problem?
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Devil’s Advocette asked:
Directions: Solve by isolationg the natural log and exponentiating both sides. Express answer in terms of e. Then use calculator to approximate solution.
Directions: Solve by isolationg the natural log and exponentiating both sides. Express answer in terms of e. Then use calculator to approximate solution.
ln(x+1) - lnx = 1
I know how to do these when it’s one natural log, but am confused on how to do it with two of them. Do log properties apply to natural logs?
The answer is supposed to come out to 1/(e-1) = aprox. 0.58
Please show me HOW to do this, thank you.
kept getting screwy answers lol, thanks for the help
Rebecca
November 18th, 2007 at 4:12 pm
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
November 18th, 2007 at 9:52 pm
yes usual log properties apply whatever the base
(x+1)/x = e
x = 1/(e-1)
November 21st, 2007 at 3:41 pm
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)
November 22nd, 2007 at 10:45 pm
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