A new car is purchased for 24900 dollars. The value of the car depreciates at 6.75% per year. To the nearest tenth of a year, how long will it be until the value of the car is 15900 dollars?

x≈6.4

Step-by-step explanation:

Exponential Functions:

y=ab^x

y=ab  

x

a=\text{starting value = }24900

a=starting value = 24900

r=\text{rate = }6.75\% = 0.0675

r=rate = 6.75%=0.0675

\text{Exponential Decay:}

Exponential Decay:

b=1-r=1-0.0675=0.9325

b=1−r=1−0.0675=0.9325

\text{Write Exponential Function:}

Write Exponential Function:

y=24900(0.9325)^x

y=24900(0.9325)  

x

Put it all together

\text{Plug in y-value:}

Plug in y-value:

15900=24900(0.9325)^x

15900=24900(0.9325)  

x

\frac{15900}{24900}=\frac{24900(0.9325)^x}{24900}

24900

15900

=  

24900

24900(0.9325)  

x

Divide both sides by 24900

0.638554=0.9325^x

0.638554=0.9325  

x

\log 0.638554=\log 0.9325^x

log0.638554=log0.9325  

x

Take the log of both sides

\log 0.638554=x\log 0.9325

log0.638554=xlog0.9325

use power rule to bring x to the front

\frac{\log 0.638554}{\log 0.9325}=\frac{x\log 0.9325}{\log 0.9325}

log0.9325

log0.638554

=  

log0.9325

xlog0.9325

Divide both sides by log(0.9325)

6.418279=x

6.418279=x

51840000 cm 2

step-by-step explanation:

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.