# Math Help - logarithmic equations

1. ## logarithmic equations

log(x-1) = 2 - log (x+1)

okay so heres what i did:
added log (x+1) to the other side, hence log (x-1) + log (x+1) = 2
then i used product rule . . .log (x-1)(x+1) = 2
now i end up getting (x+1)(x-1) = 100 after getting rid of the log with base 10, and 10^2.
x^2 = 101

idk this just seems wrong and it is probably, so if anyone can help me out i'd appreciate that a lot. thanks.

2. Hi mate

Originally Posted by rafaeli

log(x-1) = 2 - log (x+1)

This is very simple [edit: wasn't that simple, i thought log = ln,...]

log(x-1) = 2 - log (x+1)

log(x-1) + log (x+1) = 2

log([x-1][x+1]) = 2

log(x^2 - 1) = 2

//e^()

e^(log(x^2-1)) = e^2

x^2 -1 = e^2

x^2 = e^(2) +1

$x = \sqrt{e^2 +1}$

do you understand?

3. thank you so much. i understand just about all of it. except for where the e comes in. like why use e, the natural log? and why does e^log() cancel out?

4. Originally Posted by rafaeli
thank you so much. i understand just about all of it. except for where the e comes in. like why use e, the natural log? and why does e^log() cancel out?

ooops, im very sorry, i thought log =: log_e

[quote]
after getting rid of the log with base 10
[quote]

your solution x^2 = 101 is correct

what i meant was

log(x-1) = 2 - log (x+1)

log(x-1) + log (x+1) = 2

log([x-1][x+1]) = 2

log(x^2 - 1) = 2

//log_b (a) = ln(a) / ln(b)

ln(x^2 - 1)/ln(10) = 2

ln(x^2-1) = 2ln(10)

//e^()

x^2-1 = e^(2ln(10))

$x = \sqrt{e^{2ln(10)}+1} = \sqrt{101}$

Sorry, my mistake :-(

5. oh its okay, thanks for your help. the original answer i got just didn't seem right, but thanks for verifying it.