# Thread:

1. ## A Few exponential problems.

just need to get these questions out of the way, help is greatly appreciated.

first question.
The population of insects has been increasing exponentially by 15% per year how long will it take for the population to double.

*I take it i will need to create formule of something like y = (1.15)^x but how would i know what to set my y to as i am not given any initial population

Second Question

a 50mg sample of cobalt 60 decays to 40mg after 1.6 minutes how long will it take the sample to decay to 5% of its initial amount

Thanks for taking time to look at my questions

2. ## Re: A Few exponential problems.

Originally Posted by waleedrabbani
just need to get these questions out of the way, help is greatly appreciated.

first question.
The population of insects has been increasing exponentially by 15% per year how long will it take for the population to double.

*I take it i will need to create formule of something like y = (1.15)^x but how would i know what to set my y to as i am not given any initial population

let P = initial population ...

2P = P(1.15)t

Second Question

a 50mg sample of cobalt 60 decays to 40mg after 1.6 minutes how long will it take the sample to decay to 5% of its initial amount

40 = 50 e1.6k

solve for the decay constant, k , then sub in for k in the equation below ... solve for t

2.5 = 50ekt

...

3. ## Re: A Few exponential problems.

Did you mean to put e in as a variable or does it have the numerical value of 2.(digits i am unsure of)

4. ## Re: A Few exponential problems.

radioactive substances decay at a continuous exponential rate. you should already know that the equation for such decay (or growth) is

$y = y_0 e^{kt}$ where $e \approx 2.718 ...$

... use the "e" key on your calculator.

5. ## Re: A Few exponential problems.

No, e is not a variable, it is the base of the natural logarithms, about 2.7182....

You can do the second problem without using "e" (because all exponentials are equivalent). If "50mg sample of cobalt 60 decays to 40mg after 1.6 minutes" then whatever the inital amount is is multiplied by $\frac{40}{50}= \frac{4}{5}$ every 1.6 minutes. In t minutes, there are $\frac{t}{1.6}$ intervals of 1.6 minutes so after t minutes, the original amount will have multiplied by $\frac{4}{5}$ $\frac{t}{1.6}$ times: $\left(\frac{4}{5}\right)^\frac{t}{1.6}$. Set that equal to .05 and solve for t. Of course, you will need to use a logarithm to do that: $log(\left(\frac{4}{5}\right)^\frac{t}{1.6})= \frac{t}{1.6}log(\frac{4}{5})$.

"All exponential are equivalent" because $e^x$ and $ln(x)$ are inverse functions: $e^{ln(x)}= x$. In particular, $\left(\frac{4}{5}\right)^{t/1.6}=e^{ln((\frac{4}{5})^{t/1.6})= e^{(t/1.6)ln(4/5)}$
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