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Thread: Word Problem Exponential Equations /Value

  1. #1
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    Word Problem Exponential Equations /Value

    Painting that was purchased 20 years ago for $5000 is now worth $23,305. Calculate how much the painting will be worth 30 years from now?
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    Re: Word Problem Exponential Equations /Value

    Quote Originally Posted by tdotodot View Post
    Painting that was purchased 20 years ago for \$5000 is now worth \$23,305. Calculate how much the painting will be worth 30 years from now?
    V = V_0e^{kt}

    23305 = 5000e^{20k}

    solve for $k$, then use that value for $k$ to determine the value when $t = 50$
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    Re: Word Problem Exponential Equations /Value

    sorry this is my sisters high school question I'm trying to help her with.
    how do i solve for k, without e?
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    Re: Word Problem Exponential Equations /Value

    e is Euler's constant , e=2.718. The symbol 'e' should be somewhere on your calculator
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    Re: Word Problem Exponential Equations /Value

    Quote Originally Posted by tdotodot View Post
    sorry this is my sisters high school question I'm trying to help her with.
    how do i solve for k, without e?
    if that is the case, your sister should know something about using logarithms to solve for $k$ in an exponential equation
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    Re: Word Problem Exponential Equations /Value

    k=9.62698x10^-9 ?
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    Re: Word Problem Exponential Equations /Value

    Quote Originally Posted by tdotodot View Post
    k=9.62698x10^-9 ?
    not even close
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    Re: Word Problem Exponential Equations /Value

    why don't you have your sister show how she solved for k?

    can't fix it if I don't know where the mistake(s) are ...
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    Re: Word Problem Exponential Equations /Value

    Quote Originally Posted by tdotodot View Post
    sorry this is my sisters high school question I'm trying to help her with.
    how do i solve for k, without e?
    Well, if e doesn't enter the picture, then it's regular compounding.
    Tried annual : 5000(1 + i)^20 = 23305
    Solve for i : you'll get .080000496512...
    So "seems" teacher set that up so you'll get 8% annual compounding.

    If Sis can't wrap up with that, well....
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    Re: Word Problem Exponential Equations /Value

    Quote Originally Posted by skeeter View Post
    why don't you have your sister show how she solved for k?

    can't fix it if I don't know where the mistake(s) are ...
    She came to me because she did not get the question. It's been a good five years since I've seen this stuff.

    Quote Originally Posted by DenisB View Post
    Well, if e doesn't enter the picture, then it's regular compounding.
    Tried annual : 5000(1 + i)^20 = 23305
    Solve for i : you'll get .080000496512...
    So "seems" teacher set that up so you'll get 8% annual compounding.

    If Sis can't wrap up with that, well....
    That would be getting into finance, i asked her and thats her next subject. Is there not a generic formula which can be used to directly solve this type of problem?


    I know people don't like to give answers out and like to let the posters think for themselves... but I'm asking for someone to explain this to me, so i can explain it to her.
    Last edited by tdotodot; Nov 25th 2015 at 07:32 PM.
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    Re: Word Problem Exponential Equations /Value

    I provided the general formula for continuous exponential growth in post #2. If this is the formula she is to use, then she should be familiar with it. So, the question is, what has she learned about exponential growth? Kind of difficult to proceed without some additional background information. Does she recall using any formulas in class?
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    Re: Word Problem Exponential Equations /Value

    If it's the original case, to solve for k, we take the natural logarithm of each side of the equation to get:

    ln(23305) = ln(5000e^{20k})

    Using the properties of logarithms, when multiplication of two arguments is the same as adding the ln of each argument.

    ln(23305) = ln(5000) + ln(e^{20k})

    Now, we can simplify Ln(e^{20k}) by bring down the exponent and multiply it by ln (e). What is Ln(e) = ???
    Spoiler:
    ln(e)=1


    ln(23305) = ln(5000) + (20k)ln(e)


    ln(23305) = ln(5000) + (20k) \implies k = \frac{ ln(23305) - ln(5000)}{ 20} = 0.07696
    Last edited by sakonpure6; Nov 25th 2015 at 07:49 PM.
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    Re: Word Problem Exponential Equations /Value

    Well, because of the "nice" 8% result from assuming annual compounding,
    I'm bettin' ya'll a Canadian Toonie (thassa 2 dollar coin worth ~1.52US)
    that this is what the teacher expects.

    And looks like the "sister" routine is a ploy to convince us to give you full solution.
    If not, then your sister was looking at the muscular boy in the seat next to hers
    when the teacher was teaching....
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    Re: Word Problem Exponential Equations /Value

    Quote Originally Posted by tdotodot View Post
    Painting that was purchased 20 years ago for $5000 is now worth $23,305. Calculate how much the painting will be worth 30 years from now?
    If every year the value of the painting increases by a certain rate "r", then every year:

    $\displaystyle \begin{align*} V_1 &= V_0 + r\,V_0 \\ &= V_0 \left( 1 + r \right) \\ \\ V_2 &= V_1 + r\,V_1 \\ &= V_1 \left( 1 + r \right) \\ &= V_0 \left( 1 + r \right) \left( 1 + r \right) \\ &= V_0 \left( 1 + r \right) ^2 \\ \\ V_3 &= V_2 + r\,V_2 \\ &= V_2 \left( 1 + r \right) \\ &= V_0 \left( 1 + r \right) ^2 \left( 1 + r \right) \\ &= V_0 \left( 1 + r \right) ^3 \\ \\ V_n &= V_0 \left( 1 + r \right) ^n \end{align*}$

    So from this we can find

    $\displaystyle \begin{align*} 23\,305 &= 5000 \left( 1 + r \right) ^{20} \\ \frac{23\,305}{5000} &= \left( 1 + r \right) ^{20} \\ \frac{4661}{1000} &= \left( 1 + r \right) ^{20} \\ \left( \frac{4661}{1000} \right) ^{\frac{1}{20}} &= 1 + r \end{align*}$

    In 30 years it will be $\displaystyle \begin{align*} V_{50} \end{align*}$, (30 years after the 20th year) so that means

    $\displaystyle \begin{align*} V_{50} &= 5000 \left[ \left( \frac{4661}{1000} \right) ^{\frac{1}{20}} \right] ^{50} \\ &= 5000 \left( \frac{4661}{1000} \right) ^{\frac{5}{2}} \\ &\approx 234\,513.45 \end{align*}$
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    Re: Word Problem Exponential Equations /Value

    Quote Originally Posted by Prove It View Post
    In 30 years it will be $\displaystyle \begin{align*} V_{50} \end{align*}$, (30 years after the 20th year) so that means

    $\displaystyle \begin{align*} V_{50} &= 5000 \left[ \left( \frac{4661}{1000} \right) ^{\frac{1}{20}} \right] ^{50} \\ &= 5000 \left( \frac{4661}{1000} \right) ^{\frac{5}{2}} \\ &\approx 234\,513.45 \end{align*}$
    ...which is 8% cpd annually....to be precise: 8.00004965121881505....
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