I'm having difficulties solving the following equation:
(logx)+1 = log(x+1)
Any help would be greatly appreciated. Thanks.
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.
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....