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### Student Question

Secondary V • 1yr.

For the problem on top, I don't know how to do c and d. For the one down, I do not know how to even begin.

Mathematics

## Explanations (1)

• Explanation from Alloprof

Explanation from Alloprof

This Explanation was submitted by a member of the Alloprof team.

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Team Alloprof • 1yr.

Hi !

Let's begin with c :

25000 can be written as 5*5*1000

So log(25000) = log (5*5*1000)

Now you can use one of the laws of logs. In this exercise will use :

So log (5*5*1000) = log(5) + log(5) + log(1000)

Well, we know that log(1000) = 3

So we can rewrite our equation :

log (25000) = 2*log(5) + 3 = 2*v+3

I did all the c with you, but for the d I ll just give you some tips.

tip 1 : rewrite sqrt(1000) as 1000^1/2

tip 2 : use one of the laws of logs

----------------------------------------------------------------

You have to know that half-life of an isotope means the time taken for the radioactivity of a specified isotope to fall to half its original value.

You have the function :

P(t) = e^(-0.00001*t) and I suppose that p(t) is a proportion.

You have to find the time taken of your isotope to fall to half its original value, so 50%

In other words, you have to isolate t in : P(t) = e^(-0.00001*t)

Have a nice day

KH