40 PTS! EXPONENTIAL FUNCTIONS!

Some years ago, Mr. Smith & Mr Jones bought a car on the same day. The value of both cars has decreased every year.

Mr Smith: Function f can be used to determine the value of Mr. Smith's car:
f(x) = 23, 000b ˣ
Where x is the time elapsed, in years since the car was bought
& y the value of Mr. Smith's car.

Mr. Jones: Function g can be used to determine the value of Mr. Jones' car:
g(x) = a(0.82) ˣ
Where x is the time elapsed in years since the car was bought
& y the value of Mr. Jones' car.

Two years after Mr. Smith bought his car, it was valued at $14, 720.
Three years after Mr. Jones bought his car, it was valued at $16, 541. 04.
Today, the value of Mr Smith's car is $4,823.45.
What is the value of Mr Jones' car today?

Question

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Respuesta :

joshlind3 joshlind3 · Answer 1 · 2019-09-15T23:35:56+02:00

Answer:

Jones' car is worth $7478.56 now

Step-by-step explanation:

Mr Smith: [tex]f(x)=23000*b^x[/tex]

Mr Jones: [tex]g(x) = a(0.82)^x[/tex]

After two years, Mr. Smith's car is $14,720

We can use this to solve for b in his equation.

[tex]23000b^2=14720\\b^2=\frac{16}{25}\\\sqrt{b^2}=\sqrt{\frac{16}{25}} \\b = \frac{4}{5}[/tex]

Mr Smith: [tex]f(x) = 23000(\frac{4}{5})^x[/tex]

Then, we can use the value of Mr. Smith's car today to find the amount of time has passed.

[tex]23000(\frac{4}{5})^x =4823.45\\\frac{4}{5}^x=0.20971\\ ln(0.8^x)=ln(0.20971)\\ xln(0.8)=ln(0.20971)\\x=\frac{ln(0.20971)}{ln(0.8)} = 7[/tex]

For the final price, 7 years has passed.

We can use Jones' info to find his equation

[tex]a(0.82)^3=16541.04\\a(0.5513)=16541.04\\a=\frac{16541.04}{0.5513} \\a = 30000[/tex]

Jones: [tex]g(x) = 30000(0.82)^x[/tex]

Now we just plug in x = 7 to find our value

[tex]30000(0.82)^7=7478.56[/tex]