# Log

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

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

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

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

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

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

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

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

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

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

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

hmmm, I get a different answer than you do

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle tlog 12.5=log 250$

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

much better