Chapter 3.4, Problem 82E

a .

To determine

To find: the demand x for price $p=\$169$ .

Answer to Problem 82E

983 Units

Explanation of Solution

Given:

The demand equation for a Smartphone is:

  $p=5000\left(1-\frac{4}{4+e^{-0.002x}}\right)$ ….. (1)

Concept used:

Power property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a{u}^n=n{\log }_{a}u$

Substitute $p=\$169$ in equation (1),

  $\begin{array}{c}169=5000\left(1-\frac{4}{4+e^{-0.002x}}\right)\\ \frac{169}{5000}=1-\frac{4}{4+e^{-0.002x}}\\ \frac{4}{4+e^{-0.002x}}=1-\frac{169}{5000}\\ \frac{4}{4+e^{-0.002x}}=\frac{4831}{5000}\end{array}$

  $\begin{array}{c}4831\left(4+e^{-0.002x}\right)=20,000\\ 4+e^{-0.002x}=\frac{20000}{4831}\\ e^{-0.002x}=\frac{20000}{4831}-4\\ e^{-0.002x}=\frac{676}{4831}\end{array}$

Take natural logarithms on both sides,

  $\ln e^{-0.002x}=\ln \frac{676}{4831}$

Using power rule,

  $-0.002x\ln e=\ln \frac{676}{4831}$

Use $\ln e=1$ ,

  $\begin{array}{c}-0.002x\left(1\right)=\ln \frac{676}{4831}\\ -0.002x=\ln \frac{676}{4831}\\ x=\frac{\ln \frac{676}{4831}}{-0.002}\\ x\approx 983.308\end{array}$

Hence, the demand is approximately 983 units.

b .

To determine

To find: the demand x for price $p=\$299$ .

Answer to Problem 82E

684 Units

Explanation of Solution

Given:

The demand equation for a Smartphone is:

  $p=5000\left(1-\frac{4}{4+e^{-0.002x}}\right)$ ….. (1)

Concept used:

Power property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a{u}^n=n{\log }_{a}u$

Substitute $p=\$299$ in equation (1),

  $\begin{array}{c}299=5000\left(1-\frac{4}{4+e^{-0.002x}}\right)\\ \frac{299}{5000}=1-\frac{4}{4+e^{-0.002x}}\\ \frac{4}{4+e^{-0.002x}}=1-\frac{299}{5000}\\ \frac{4}{4+e^{-0.002x}}=\frac{4701}{5000}\end{array}$

  $\begin{array}{c}4701\left(4+e^{-0.002x}\right)=20,000\\ 4+e^{-0.002x}=\frac{20000}{4701}\\ e^{-0.002x}=\frac{20000}{4701}-4\\ e^{-0.002x}=\frac{1196}{4701}\end{array}$

Take natural logarithms on both sides,

  $\ln e^{-0.002x}=\ln \frac{1196}{4701}$

Using power rule,

  $-0.002x\ln e=\ln \frac{1196}{4701}$

Use $\ln e=1$ ,

  $\begin{array}{c}-0.002x\left(1\right)=\ln \frac{1196}{4701}\\ -0.002x=\ln \frac{1196}{4701}\\ x=\frac{\ln \frac{1196}{4701}}{-0.002}\\ x\approx 684.396\end{array}$

Hence, the demand is approximately 684 units.