# Math Help - y=abx formula help

1. ## 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.

2. Originally Posted by azncocoluver
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.

3. ## 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.

4. Originally Posted by earboth
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.

5. Originally Posted by MathQueen
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

$
500=2200\cdot (0.82)^x
$

for x:

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

Therefore

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