# y=abx formula help

• Nov 24th 2008, 06:13 PM
azncocoluver
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.
• Nov 24th 2008, 10:24 PM
earboth
Quote:

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:

$\displaystyle 500=2200\cdot (0.88)^x$

3. For confrimation: I've got 11.6 years.
• Apr 8th 2009, 02:11 PM
MathQueen
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: $\displaystyle 500=2200\cdot (0.82)^x$. 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.
• Apr 8th 2009, 02:13 PM
MathQueen
Quote:

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:

$\displaystyle 500=2200\cdot (0.88)^x$

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.
• Apr 9th 2009, 12:07 AM
earboth
Quote:

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

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

for x:

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

Therefore

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