Thread:

1. 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.
$1000(1.10)^x=2000(0.95)^x$
Then by definition $C^x=y <==> log(y)=x$

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

And now I'm stuck, help!

2. 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.
$1000(1.10)^x=2000(0.95)^x$
To make things easier, divide both sides by 1000.

What next?

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

What next?
$(1.1)^x=2(0.95)^x$

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

?

4. Using what definition? Taking the logarithm of both sides (any base) of
$(1.1)^x= 2(0.95)^x$ you get

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