# Help with log problem

• Aug 25th 2007, 12:40 AM
gregorio
Help with log problem
I am not sure what to use for this problem, I think it is a log is there any formula I could use to solve this ? Any help appreciated, I am studying problems from old exams to practicing for my exam !!

A ship is leaking oil, it doubles the area it covers every 7 days. If the initial leakage area is 15 square metres how long would it take to cover 25000 metres squared.
• Aug 25th 2007, 01:07 AM
rualin
I had typed out a nice explanation but firefox had an error so here it goes again. I have learned to copy and temporarily save as I type now. :(

It seems to be an exponential function since it doubles every week, $x$. So $25000 = 15\cdot 2^x$

Dividing by 15, we get $\frac{5000}{3} = 2^x$

Taking the log base two of each side, we get $\log_2{\frac{5000}{3}} = \log_2{2^x} = x\cdot \log_2{2} = x$

So $\boxed{x \approx 10.71~ weeks}$ or if needed in days, multiply by 7, of course.
• Aug 25th 2007, 01:22 AM
CaptainBlack
Quote:

Originally Posted by gregorio
I am not sure what to use for this problem, I think it is a log is there any formula I could use to solve this ? Any help appreciated, I am studying problems from old exams to practicing for my exam !!

A ship is leaking oil, it doubles the area it covers every 7 days. If the initial leakage area is 15 square metres how long would it take to cover 25000 metres squared.

After 7 days the covered area is 2 x 15 m^2
After 14 days the covered area is 4 x 15 m^2
After 21 days the covered area is 2^3 x 15 m^2

After 7N days the covered area is 2^N x 15 m^2

If after 7N days the area covered is 25000 m^2, we have:

25000 = 2^N x 15.

Now take logs (any base it does not matter):

log(25000) = N log(2) + log(15),

so:

N = [log(25000) - log(15)]/log(2) = log(5000/3)/log(2) ~=10.7

So the oil covers the required area after about 7 x 10.7 ~= 75 days

RonL