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Math Help - Exponential Equation & Decay

  1. #1
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    Question Exponential Equation & Decay

    Hello there! I was just wondering if someone could kindly show me how to do the following questions:

    1) 4^x + 6(4^-x) = 5

    2) The half-life of a certain substance is 3.6 days. How long will it take for 20g of the substance to decay to 7g?
    -The only equation I know of is y = ab^x but I'm not sure what to put where, thank you!!
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  2. #2
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    Hint: use logarithms: log (x^y) = y log x
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  3. #3
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    Quote Originally Posted by Marmalade View Post
    Hello there! I was just wondering if someone could kindly show me how to do the following questions:

    1) 4^x + 6(4^-x) = 5

    2) The half-life of a certain substance is 3.6 days. How long will it take for 20g of the substance to decay to 7g?
    -The only equation I know of is y = ab^x but I'm not sure what to put where, thank you!!
    4^x + 6(4^{-x}) = 5

    multiply every term by 4^x ...

    4^{2x} + 6 = 5(4^x)

    4^{2x} - 5(4^x) + 6 = 0

    (4^x - 3)(4^x - 2) = 0

    4^x = 3 ... 4^x = 2

    x = \frac{\log{3}}{\log{4}} ... x = \frac{1}{2}



    y = 20\left(\frac{1}{2}\right)^{\frac{t}{3.6}}<br />

    set y = 7 and solve for t
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  4. #4
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    Hello, Marmalade!

    2) The half-life of a certain substance is 3.6 days.
    How long will it take for 20g of the substance to decay to 7g?

    The only equation I know of is: . y \:=\: ab^x
    but I'm not sure what to put where, thank you!
    You can derive the half-life formula yourself . . .

    The general form is: . A \:=\:A_oe^{-kt} .[1]

    . . where: . \begin{array}{ccc}A &=& \text{amount at time }t \\ A_o &=& \text{original amount} \\ k &=& \text{constant} \end{array}


    We are given: . \text{amount }\tfrac{1}{2}A_o\text{ at }t = 3.6

    Substitute into [1]: . \tfrac{1}{2}A_o \:=\:A_oe^{-3.6k} \quad\Rightarrow\quad e^{-3.6k} \:=\:\frac{1}{2}

    Take logs: . \ln\left(e^{-3.6k}\right) \:=\:\ln\left(\tfrac{1}{2}\right) \:=\:\ln\left(2^{-1}\right) \:=\:-\ln2

    . . Then: . -3.6k \:=\:-\ln 2 \quad\Rightarrow\quad k \:=\:\frac{\ln2}{3.6}


    Hence, the function is: . A \:=\:A_oe^{-\frac{\ln2}{3.6}t}

    For this problem, A_o = 20\quad\Rightarrow\quad A \:=\:20e^{-\frac{\ln2}{3.6}t}



    Now we can answer the question: When will A = 7 ?

    We have: . 7 \:=\:20e^{-\frac{\ln2}{3.6}t} \quad\Rightarrow\quad e^{-\frac{\ln2}{3.6}t} \:=\:0.35 \quad\Rightarrow\quad -\frac{\ln2}{3.6}t \:=\:\ln(0.35)

    . . t \:=\:-\frac{3.6}{\ln2}\!\cdot\!\ln(0.35) \:=\:5.452463422 \;\approx\;5.45 days

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Marmalade!

    You can derive the half-life formula yourself . . .

    The general form is: . A \:=\:A_oe^{-kt} .[1]

    . . where: . \begin{array}{ccc}A &=& \text{amount at time }t \\ A_o &=& \text{original amount} \\ k &=& \text{constant} \end{array}


    We are given: . \text{amount }\tfrac{1}{2}A_o\text{ at }t = 3.6


    Substitute into [1]: . \tfrac{1}{2}A_o \:=\:A_oe^{-3.6k} \quad\Rightarrow\quad e^{-3.6k} \:=\:\frac{1}{2}

    Take logs: . \ln\left(e^{-3.6k}\right) \:=\:\ln\left(\tfrac{1}{2}\right) \:=\:\ln\left(2^{-1}\right) \:=\:-\ln2

    . . Then: . -3.6k \:=\:-\ln 2 \quad\Rightarrow\quad k \:=\:\frac{\ln2}{3.6}


    Hence, the function is: . A \:=\:A_oe^{-\frac{\ln2}{3.6}t}

    For this problem, A_o = 20\quad\Rightarrow\quad A \:=\:20e^{-\frac{\ln2}{3.6}t}



    Now we can answer the question: When will A = 7 ?

    We have: . 7 \:=\:20e^{-\frac{\ln2}{3.6}t} \quad\Rightarrow\quad e^{-\frac{\ln2}{3.6}t} \:=\:0.35 \quad\Rightarrow\quad -\frac{\ln2}{3.6}t \:=\:\ln(0.35)

    . . t \:=\:-\frac{3.6}{\ln2}\!\cdot\!\ln(0.35) \:=\:5.452463422 \;\approx\;5.45 days
    Hello Sorobon! Wow, that certainly is a new equation for me. Thank you for the step by step instructions, I should be able to get it!
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  6. #6
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    Quote Originally Posted by Marmalade View Post
    2) The half-life of a certain substance is 3.6 days. How long will it take for 20g of the substance to decay to 7g?
    -The only equation I know of is y = ab^x but I'm not sure what to put where, thank you!!
    Another way to do this is to recognize that "half" life means that the amount is multiplied by 1/2 every 3.6 days- In T days, there are T/3.6 periods of 3.6 days each which means the initial amount is multiplied by 1/2 T/3.6 times: x= 20(1/2)^{T/3.6}. You want to solve 7= 20(1/2)^{T/3.6} or, dividing both sides by 20, \frac{7}{20}= (1/2)^{T/3.6}. To solve that take the logarithm of both sides (it doesn't matter whether you take the natural logarithm or common logarithm) log(\frac{7}{20})= (T/3.6)log(1/2) so [tex]T= 3.6\frac{log(7/20)}{log(1/2)}= 3.6\frac{log(20)- log(7)}{log(2)}, exactly what Soroban got.

    The moral is that all logarithms and all exponentials are equivalent. It doesn't matter what base you use so use whatever is simplest.
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Another way to do this is to recognize that "half" life means that the amount is multiplied by 1/2 every 3.6 days- In T days, there are T/3.6 periods of 3.6 days each which means the initial amount is multiplied by 1/2 T/3.6 times: x= 20(1/2)^{T/3.6}. You want to solve 7= 20(1/2)^{T/3.6} or, dividing both sides by 20, \frac{7}{20}= (1/2)^{T/3.6}. To solve that take the logarithm of both sides (it doesn't matter whether you take the natural logarithm or common logarithm) log(\frac{7}{20})= (T/3.6)log(1/2) so [tex]T= 3.6\frac{log(7/20)}{log(1/2)}= 3.6\frac{log(20)- log(7)}{log(2)}, exactly what Soroban got.

    The moral is that all logarithms and all exponentials are equivalent. It doesn't matter what base you use so use whatever is simplest.
    Oh, thank you HallsofIvy, this is actually a lot simpler to work with! I appreciate it.
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