I'm having difficulties solving the following equation:

(logx)+1 = log(x+1)

Any help would be greatly appreciated. Thanks.

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- Apr 8th 2009, 12:15 PMSodapopLogarithm problem
I'm having difficulties solving the following equation:

(logx)+1 = log(x+1)

Any help would be greatly appreciated. Thanks. - Apr 8th 2009, 12:17 PMJhevon
- Apr 8th 2009, 02:31 PMSodapop
Hmm, then...

$\displaystyle \log \frac {x+1}{x} = 1$

$\displaystyle 10^1 = \frac {x+1}{x}$

$\displaystyle 10x = x+1$

$\displaystyle 9x = 1$

$\displaystyle x = \frac 19$

Oh, awesome. Thanks, I simply haven't thought about transforming the log into a $\displaystyle 10^x$ form. You're the best. (Clapping) - Apr 8th 2009, 02:50 PMJhevon
- Apr 8th 2009, 02:54 PMpickslides
Hey there Sodapop

I would've made

$\displaystyle

\log x + 1 = \log (x + 1) $

into

$\displaystyle

\log x + \log 10 = \log (x + 1) $

assuming we are in base ten.

then using the rule for addition of logs we get,

$\displaystyle

\log 10x = \log (x + 1) $

removing the log from both sides we get

$\displaystyle

10x = x + 1

$

and thus getting

$\displaystyle x = \frac{1}{9}$

More than one way to skin a cat!

Sorry to all those cat lovers....