# Thread: logarithims

1. ## logarithims

i'm getting stuck ....a little help please

$5^{x-2} = 4^x$

i got..
$log 5 ^{x-2} = log 4^x$
$(x-2) log 5 = x log 4$

and i'm stuck here...

2. Originally Posted by extraordinarymachine
i'm getting stuck ....a little help please

$5^x-2 = 4^x$

i got..
$log 5 ^x-2 = log 4^x$ <<<<<<<<< never commit such a crime again!
$(x-2) log 5 = x log 4$

and i'm stuck here...
Do you mean

$5^x - 2 = 4^x$ If so

$\left(\dfrac54 \right)^x = 2~\implies~x=\log_{\frac54}(2)=\dfrac{\ln(2)}{\ln( 5)-\ln(4)}$

or

$5^{x-2} = 4^x$ If so

$\frac1{25} \cdot 5^x = 4^x~\implies~\left(\dfrac54 \right)^x = 25 ~\implies~x=\log_{\frac54}(25)=\dfrac{\ln(25)}{\ln (5)-\ln(4)}$

3. Originally Posted by earboth
Do you mean

$5^x - 2 = 4^x$ If so

sorry i messed up the question, i ment to write it like this

4. Originally Posted by extraordinarymachine
i'm getting stuck ....a little help please

$5^{x-2} = 4^x$

i got..
$log 5 ^{x-2} = log 4^x$
$(x-2) log 5 = x log 4$

and i'm stuck here...
$(x-2) log 5 = x log 4
\Leftrightarrow \frac{ x-2}{x }$
$= \frac{log4 }{log5 } \Leftrightarrow 1- \frac{ 2}{x } =\frac{log4 }{log5 }$

5. $(x-2)log5=xlog4$

$\frac{x-2}{x}=\frac{log4}{log2}$

$1-\frac{2}{x}=\frac{log4}{log5}$

$\frac{2}{x}=1-\frac{log4}{log5}$

$\frac{2}{x}=\frac{log5}{log5}-\frac{log4}{log5}$

$\frac{2}{x}=\frac{log5-log4}{log5}$

$\frac{x}{2}=\frac{log5}{log5-log4}$

$x=\frac{2log5}{log5-log4}$

6. 5 steps
$(x-2)log5=xlog4$
$xlog5-2log5=xlog4$
$xlog5-xlog4=2log5$
$x(log5-log4)=2log5$
$x=\frac{2log5}{log5-log4}$

7. Originally Posted by Krahl
5 steps
$(x-2)log5=xlog4$
$xlog5-2log5=xlog4$
$xlog5-xlog4=2log5$
$x(log5-log4)=2log5$
$x=\frac{2log5}{log5-log4}$
this was really helpful thanks!