# Logarithm problem

• Apr 8th 2009, 12:15 PM
Sodapop
Logarithm 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 PM
Jhevon
Quote:

Originally Posted by Sodapop
I'm having difficulties solving the following equation:

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

Any help would be greatly appreciated. Thanks.

note that you have: $\displaystyle 1 = \log (x + 1) - \log x$

now recall: $\displaystyle \log X - \log Y = \log \frac XY$ and $\displaystyle \log_a b = c \implies a^c = b$
• Apr 8th 2009, 02:31 PM
Sodapop
Quote:

Originally Posted by Jhevon
note that you have: $\displaystyle 1 = \log (x + 1) - \log x$

now recall: $\displaystyle \log X - \log Y = \log \frac XY$ and $\displaystyle \log_a b = c \implies a^c = b$

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 PM
Jhevon
Quote:

Originally Posted by Sodapop
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)

yup, that's it. good job
• Apr 8th 2009, 02:54 PM
pickslides
Hey there Sodapop

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

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....