# Logarithm problem

• Apr 8th 2009, 01:15 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, 01:17 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: $1 = \log (x + 1) - \log x$

now recall: $\log X - \log Y = \log \frac XY$ and $\log_a b = c \implies a^c = b$
• Apr 8th 2009, 03:31 PM
Sodapop
Quote:

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

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

Hmm, then...

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

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

$10x = x+1$

$9x = 1$

$x = \frac 19$

Oh, awesome. Thanks, I simply haven't thought about transforming the log into a $10^x$ form. You're the best. (Clapping)
• Apr 8th 2009, 03:50 PM
Jhevon
Quote:

Originally Posted by Sodapop
Hmm, then...

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

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

$10x = x+1$

$9x = 1$

$x = \frac 19$

Oh, awesome. Thanks, I simply haven't thought about transforming the log into a $10^x$ form. You're the best. (Clapping)

yup, that's it. good job
• Apr 8th 2009, 03:54 PM
pickslides
Hey there Sodapop

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

into

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

assuming we are in base ten.

then using the rule for addition of logs we get,

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

removing the log from both sides we get

$
10x = x + 1
$

and thus getting

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

More than one way to skin a cat!

Sorry to all those cat lovers....