Chapter 5.6, Problem 98AYU

(A)

To determine

To solve:

The function for $t$ in term of $V$ and the car be worth

Answer to Problem 98AYU

The car be worth is $t\approx 1.1$

Explanation of Solution

Given:

  $V=6.78{\left(1.0114\right)}^{t-2009}$

Concept used:

If $x$ and $b$ are positive real numbers and $b\ne 1$

  ${\log }_{b}x=y$ Is same value $b^y=x$

  $\log a+\log b=\log ab$

  $\log a-\log b=\log \left(\frac{a}{b}\right)$

Calculation:

In exponential form using the definition of a logarithm

Using the formula

  ${\log }_{b}x=y$ Is same value $b^y=x$

Using the formula

  $\begin{array}{l}V=6.78{\left(1.0114\right)}^{t-2009}\\ \frac{V}{6.78}={\left(1.0114\right)}^{t-2009}\\ \ln \left(\frac{V}{6.78}\right)=\ln {\left(1.0114\right)}^{t-2009}\\ \frac{\ln \left(\frac{V}{6.78}\right)}{\ln \left(1.0114\right)}=t-2009\end{array}$

The car be worth $8.7\text{million}$

  $\begin{array}{l}\frac{\ln \left(\frac{V}{6.78}\right)}{\ln \left(1.0114\right)}=t-2009\\ \frac{\ln \left(\frac{8.7,0000000}{6.78}\right)}{\ln \left(1.0114\right)}=t-2009\\ t\approx 1.1\end{array}$

(B)

To determine

To solve:

The function for $t$ in term of $V$ and the car be worth

Answer to Problem 98AYU

The car be worth is $t\approx 7.4$

Explanation of Solution

Given:

  $V=6.78{\left(1.0114\right)}^{t-2009}$

Concept used:

If $x$ and $b$ are positive real numbers and $b\ne 1$

  ${\log }_{b}x=y$ Is same value $b^y=x$

  $\log a+\log b=\log ab$

  $\log a-\log b=\log \left(\frac{a}{b}\right)$

Calculation:

In exponential form using the definition of a logarithm

Using the formula

  ${\log }_{b}x=y$ Is same value $b^y=x$

Using the formula

  $\begin{array}{l}V=6.78{\left(1.0114\right)}^{t-2009}\\ \frac{V}{6.78}={\left(1.0114\right)}^{t-2009}\\ \ln \left(\frac{V}{6.78}\right)=\ln {\left(1.0114\right)}^{t-2009}\\ \frac{\ln \left(\frac{V}{6.78}\right)}{\ln \left(1.0114\right)}=t-2009\end{array}$

The car be worth $8.7\text{million}$

  $\begin{array}{l}\frac{\ln \left(\frac{V}{6.78}\right)}{\ln \left(1.0114\right)}=t-2009\\ \frac{\ln \left(\frac{14,0000000}{6.78}\right)}{\ln \left(1.0114\right)}=t-2009\\ t\approx 7.4\end{array}$