Chapter 6.6, Problem 58E

To determine

To calculate: The number of time periods when cars price exceed.

Answer to Problem 58E

The required number of times period is $4.26$ .

Explanation of Solution

Given Information:

Purchasing amount of car $=\$20000$ .

Rate $=15\%$ (decreasing).

Also, the car value is $\$10000$ .

Formula used:

Compound interest formula:

  $A=P{\left(1+r\right)}^t$

Where

  $\begin{array}{l}A=\text{final amount}\\ P=\text{initial principal amount}\\ t=\text{time periods}\end{array}$

Calculation:

Consider purchasing amount is. $\$20000$ and rate is $15\%$ .

Rate can be rewrite as.

Rate $\begin{array}{l}=15\%\\ =\frac{15}{100}\\ =0.15\end{array}$

Since, the rates is decreased.

So, rates should be in negative.

Thus, rate is $-0.15$ .

Now, for the cars value exceed $\$10000$ .

Use compound interest formula.

  $A=P{\left(1+r\right)}^t$

Substitute $10000$ for $A$ , $20000$ for $P$ , and $-0.15$ for $r$ in the above formula.

  $10000=20000{\left\{1+\left(-0.15\right)\right\}}^t$

Simplify the above equation.

  $\begin{array}{l}10000=20000{\left\{1+\left(-0.15\right)\right\}}^t\\ \frac{10000}{20000}={\left(1-0.15\right)}^t\\ 0.5={\left(0.85\right)}^t\end{array}$

Take logarithm each side of the above equation.

  $\log \left(0.5\right)=\log {\left(0.85\right)}^t$

Use power rule of logarithm $\log x^a=a\log x$ in the above equation.

  $\log \left(0.5\right)=t\log \left(0.85\right)$

Substitute $-0.301$ for $\log \left(0.5\right)$ and $-0.0705$ for $\log \left(0.85\right)$ in the above equation.

  $-0.301=t\left(-0.0705\right)$

Simplify the above equation.

  $\begin{array}{c}\left(-0.301\right)=t\left(-0.0705\right)\\ \frac{\left(-0.301\right)}{\left(-0.0705\right)}=t\\ 4.26=t\\ t=4.26\end{array}$

Hence, the require time period are $4.26$ .