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Secondary V • 1yr.
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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
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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 :

    image.png

    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

    image.png

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

    Now for your second exercise.

    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