A limited-edition poster increases in value each year. After 1 year, the poster is worth $20.70. After 2 years, it is worth $23.81. Which equation can be used to find the value, y, after x years? (Round money values to the nearest penny.)

y = 18(1.15)x
y = 18(0.15)x
y = 20.7(1.15)x
y = 20.7(0.15)x

Question

A limited-edition poster increases in value each year. After 1 year, the poster is worth $20.70. After 2 years, it is worth $23.81. Which equation can be used to find the value, y, after x years? (Round money values to the nearest penny.)

y = 18(1.15)x
y = 18(0.15)x
y = 20.7(1.15)x
y = 20.7(0.15)x

SmokeyBrown · Answer

Try to work backwards for this problem. What is the change in value between year 1 and year 2? Divide the value of the poster after 2 years by the value after 1 year, and you should get a ratio. That will be the slope of the linear equation.
Then divide the value after one year by that ratio to get the starting value of the poster, before any time had passed.

sillybilly123 · Answer

Seems this is a case of : $y(x) = A \cdot B^x$

Because you are given multiple-choice options, demanding exponential solutions.

Try $x = 1$

$y(1) = 1.8 \cdot 1.15^1 = 20.7$

--> " After 1 year, the poster is worth $20.70"

Try Y2 for yourself

You can only draw conclusions from Y1 & Y2 by knowing the relationship in-between, which is Xponential in this example :)