# Thread: Logarithm problem

1. ## Logarithm problem

I'm having difficulties solving the following equation:

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

Any help would be greatly appreciated. Thanks.

2. 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 $\displaystyle \log_a b = c \implies a^c = b$

3. 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.

4. 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.
yup, that's it. good job

5. 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....