logarithims

**extraordinarymachine** · December 17th 2009, 02:13 AM

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...

**earboth** · December 17th 2009, 02:30 AM

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

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

**extraordinarymachine** · December 17th 2009, 02:36 AM

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

**Raoh** · December 17th 2009, 02:44 AM

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...

$(x-2) log 5 = x log 4<br /> \Leftrightarrow \frac{ x-2}{x }$ $= \frac{log4 }{log5 } \Leftrightarrow 1- \frac{ 2}{x } =\frac{log4 }{log5 }$

**Stroodle** · December 17th 2009, 02:48 AM

$(x-2)log5=xlog4$

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

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

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

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

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

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

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