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Math Help - y=abx formula help

  1. #1
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    y=abx formula help

    Can someone help me with the steps to solve this problem?

    Darius bought a flat-screen TV for $2200. The value decreases by about 18% each year. How long will it take for the value to be reduced to $500.

    My Algebra 2 teacher taught us the formula y = abx where x is an exponent, so I already set up the equation.

    500 = 2200(.18) power of x
    i got .863 as x but I don't know how long that is in time or if I even did it right.
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  2. #2
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    Quote Originally Posted by azncocoluver View Post
    Can someone help me with the steps to solve this problem?

    Darius bought a flat-screen TV for $2200. The value decreases by about 18% each year. How long will it take for the value to be reduced to $500.

    My Algebra 2 teacher taught us the formula y = abx where x is an exponent, so I already set up the equation.

    500 = 2200(.18) power of x
    i got .863 as x but I don't know how long that is in time or if I even did it right.
    1. You have to calculate the remaining value. That means: If the value decreases by 18% per year, the remaining value is 88% per year.

    2. Therefore solve for x:

    500=2200\cdot (0.88)^x

    3. For confrimation: I've got 11.6 years.
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  3. #3
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    Graphing Calculator

    I was taught to use a graphing calculator. First you have to move the decimal point two places to the left to get .18. Then you do 1-.18 to get .82. Next you do: 500=2200\cdot (0.82)^x. You plug the 500 into [tex]Y1[tex] on a graphing calculator and plug the rest into [tex]Y2[tex]. then you see when the [tex]Y1[tex] and [tex]Y2[tex] are equal.
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  4. #4
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    Quote Originally Posted by earboth View Post
    1. You have to calculate the remaining value. That means: If the value decreases by 18% per year, the remaining value is 88% per year.

    2. Therefore solve for x:

    500=2200\cdot (0.88)^x

    3. For confrimation: I've got 11.6 years.
    You did it wrong because it is not .88 it is .82.
    X is an exponent not a .88X.
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  5. #5
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    Quote Originally Posted by MathQueen View Post
    You did it wrong because it is not .88 it is .82.
    X is an exponent not a .88X.
    You are right, sorry!

    You have to solve the equation

    <br />
500=2200\cdot (0.82)^x<br />

    for x:

     500=2200\cdot (0.82)^x ~\implies~0.82^x=\dfrac{500}{2200} = \dfrac5{22}

    Therefore

    x = \log_{0.82}\left(  \dfrac5{22} \right) = \dfrac{\ln\left( \frac5{22}  \right)}{\ln(0.82)} \approx 7.5\ years
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