# Log System of equations

• Dec 9th 2010, 03:20 PM
Gerard
Log System of equations
I'm needing a little assistance with this problem, haven't any trouble until this.

A village of 1000 inhabitants increases at a rate of 10% per year. A neighbouring village of 2000 inhabitants decreases at a rate of 5% per year. After how many years will these two villages have the same population?

So I wrote them as two exponential functions and made them equal to each other. Not sure this is how I should do it but it feels right.
\$\displaystyle 1000(1.10)^x=2000(0.95)^x\$
Then by definition \$\displaystyle C^x=y <==> log(y)=x\$

Turned them into this
\$\displaystyle log(1.1)1000=log(0.95)2000\$
then I know it can be turned into this.
\$\displaystyle log1000/log1.1=log2000/log0.95\$

And now I'm stuck, help!
• Dec 9th 2010, 03:26 PM
pickslides
Quote:

Originally Posted by Gerard
So I wrote them as two exponential functions and made them equal to each other. Not sure this is how I should do it but it feels right.
\$\displaystyle 1000(1.10)^x=2000(0.95)^x\$

To make things easier, divide both sides by 1000.

What next?
• Dec 9th 2010, 03:33 PM
Gerard
Quote:

Originally Posted by pickslides
To make things easier, divide both sides by 1000.

What next?

\$\displaystyle (1.1)^x=2(0.95)^x\$

Then using the definition...
\$\displaystyle log1/log1.1 = log2/log0.95\$

?
• Dec 10th 2010, 02:01 AM
HallsofIvy
Using what definition? Taking the logarithm of both sides (any base) of
\$\displaystyle (1.1)^x= 2(0.95)^x\$ you get

x log(1.1)= x log(.95)+ log 2