Chapter 3.4, Problem 84E

a.

To determine

Find the demand of a smartphone for the given price.

Answer to Problem 84E

The demand of a smartphone for the given price $p=\$169$ is $x\approx 983$ .

Explanation of Solution

Given:

The given demand equation is $p=5000\left(1-\frac{4}{4+e^{-0.002x}}\right)$ and the given price is $p=\$169$ .

Calculation:

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

Substitute the value of $p=\$169$ in the given equation.

  $\Rightarrow 169=5000\left(1-\frac{4}{4+e^{-0.002x}}\right)$

Divide both sides by $5000$ .

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

Use cross multiplication.

  $\begin{array}{l}\Rightarrow \left(4+e^{-0.002x}\right)=5000e^{-0.002x}\\ \Rightarrow 676+169e^{-0.002x}=5000e^{-0.002x}\end{array}$

Subtract $169e^{-0.002x}$ from both sides.

  $\begin{array}{l}\Rightarrow 676+169e^{-0.002x}-169e^{-0.002x}=5000e^{-0.002x}-169e^{-0.002x}\\ \Rightarrow 676=4831e^{-0.002x}\end{array}$

Divide both sides by $4831$ .

  $\begin{array}{l}\Rightarrow \frac{676}{4831}=\frac{4831}{4831}{e}^{-0.002x}\\ \Rightarrow \frac{676}{4831}=e^{-0.002x}\\ \Rightarrow e^{-0.002x}=\frac{676}{4831}\end{array}$

Taking natural log on both sides.

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

Use power property $\ln e^x=x$

  $\begin{array}{l}\Rightarrow -0.002x=\ln \frac{676}{4831}\\ \Rightarrow -0.002x=\ln \left(676\right)-\ln \left(4831\right)\\ \Rightarrow 0.002x=\ln \left(4831\right)-\ln \left(676\right)\\ \Rightarrow 0.002x=8.48281-6.51619\\ \Rightarrow 0.002x=1.9666\end{array}$

Divide both sides by $0.0002$ .

  $\begin{array}{l}\Rightarrow \frac{0.002}{0.002}x=\frac{1.9666}{0.002}\\ \Rightarrow x\approx 983.308\end{array}$

Hence thedemand of a smartphone for the given price $p=\$169$ is $x\approx 983$ .

b.

To determine

Find the demand of a smartphone for the given price.

Answer to Problem 84E

The demand of a smartphone for the given price $p=\$299$ is $x\approx 684$ .

Explanation of Solution

Given:

The given demand equation is $p=5000\left(1-\frac{4}{4+e^{-0.002x}}\right)$ and the given price is $p=\$299$ .

Calculation:

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

Substitute the value of 13

  $p=\$299$ in the given equation.

  $\Rightarrow 299=5000\left(1-\frac{4}{4+e^{-0.002x}}\right)$

Divide both sides by $5000$ .

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

Use cross multiplication.

  $\begin{array}{l}\Rightarrow 299\left(4+e^{-0.002x}\right)=5000e^{-0.002x}\\ \Rightarrow 1196+299e^{-0.002x}=5000e^{-0.002x}\end{array}$

Subtract $299e^{-0.002x}$ from both sides.

  $\begin{array}{l}\Rightarrow 1196+299e^{-0.002x}-299e^{-0.002x}=5000e^{-0.002x}-299e^{-0.002x}\\ \Rightarrow 1196=4701e^{-0.002x}\end{array}$

Divide both sides by $4701$ .

  $\begin{array}{l}\Rightarrow \frac{1196}{4701}=\frac{4701}{4701}{e}^{-0.002x}\\ \Rightarrow \frac{1196}{4701}=e^{-0.002x}\\ \Rightarrow e^{-0.002x}=\frac{1196}{4701}\end{array}$

Taking natural log on both sides.

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

Use power property $\ln e^x=x$

  $\begin{array}{l}\Rightarrow -0.002x=\ln \frac{1196}{4701}\\ \Rightarrow -0.002x=\ln \left(1196\right)-\ln \left(4701\right)\\ \Rightarrow 0.002x=\ln \left(4701\right)-\ln \left(1196\right)\\ \Rightarrow 0.002x=8.45553-7.08674\\ \Rightarrow 0.002x=1.36879\end{array}$

Divide both sides by $0.0002$ .

  $\begin{array}{l}\Rightarrow \frac{0.002}{0.002}x=\frac{1.36879}{0.002}\\ \Rightarrow x=684.396\end{array}$

Hence the demand of a smartphone for the given price $p=\$299$ is $x\approx 684$ .