Logarithm problem

**Sodapop** · April 8th 2009, 12:15 PM

I'm having difficulties solving the following equation:

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

Any help would be greatly appreciated. Thanks.

**Jhevon** · April 8th 2009, 12:17 PM

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 $\log_a b = c \implies a^c = b$

**Sodapop** · April 8th 2009, 02:31 PM

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.

**Jhevon** · April 8th 2009, 02:50 PM

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.

yup, that's it. good job

**pickslides** · April 8th 2009, 02:54 PM

Hey there Sodapop

I would've made

$ \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!

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