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) ˣWhere 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?

Answer:Jones' car is worth $7478.56 nowStep-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 $14,720We 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]