# Log

• Dec 2nd 2009, 11:20 AM
BabyMilo
Log
Prove that if $5^{3t-2}=10^{t+1}$ then $t=\frac{log 250}{log 12.5}$

I know how to solve it. just duno how to rearrange to get answer required.

thanks
• Dec 2nd 2009, 12:20 PM
BabyMilo
Quote:

Originally Posted by RRH
$
5^{3t-2}=10^{t-1}
$

$
log 5^{3t-2}=log 10^{t-1}
$

$
(3t-2)log 5=(t-1)log 10
$

$
3tlog 5-2log5=tlog 10-log 10
$

$
3tlog 5-tlog 10=2log 5-log10
$

$
t(3log 5-log 10)=2log 5-log 10
$

$
t(log 5^3-log10)=log 5^2-log 10
$

$
t(log125-log 10)=log 25-log 10
$

$
t(\frac{log 125}{log 10})=\frac{log 25}{log 10}
$

$
t=\frac{log 25}{log 10}*\frac{log 10}{log 125}
$

hmmm, I get a different answer than you do

$
t=\frac{log 250}{log 1250}
$

sry it's plus 1 not minus 1
• Dec 2nd 2009, 12:39 PM
RRH
Quote:

Originally Posted by BabyMilo
sry it's plus 1 not minus 1

no problem I will rework it so you can see the steps

$
5^{3t-2}=10^{t+1}
$

$
log 5^{3t-2}=log 10{t+1}
$

$
(3t-2)log 5=(t+1)log 10
$

$
3tlog 5-2log 5=tlog 10+log 10
$

$
3tlog 5-tlog 10=log 10+2log 5
$

$
t(3log 5-log 10)=log 10+2log5
$

$
t(log5^3-log10)=log 10+log5^2
$

$
t(log 125-log 10)=log 10+log 25
$

$
tlog (\frac{125}{10})=log 250
$

$
tlog 12.5=log 250
$

$
t=\frac{log 250}{log 12.5}
$

much better