Chapter 7, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

1. a. The variable y = t + 1 doubles in value whenever t increases by 1 unit.
2. b. The function y = Ae^0.1t increases by 10% when t increases by 1 unit.
3. c. ln xy = (ln x)(ln y).
4. d. $\sinh (\ln x)=\frac{x^2-1}{2x}.$

To determine

To explain: Whether the statement “The variable $y=t+1$ doubles in value whenever t increases by 1 unit” is true or false.

Explanation of Solution

Suppose $t=2$, the value of y becomes,

$\begin{array}{c}y=2+1\\ =3\end{array}$

If t is increased by 1 unit, $t=3$,

$\begin{array}{c}y=2+1\\ =3\end{array}$

Hence, the value of y will not be doubled whenever t increases by 1 unit.

Therefore, the given statement is false.

To determine

To explain: Whether the statement “The function $y=A{e}^{0.1t}$ increases by 10% when t increases by 1 unit” is true or false.

Explanation of Solution

Suppose $t=2$, then

$\begin{array}{c}{y}_1=A{e}^{0.1\times 2}\\ =A{e}^{0.2}\\ \approx 1.221A\end{array}$

If t is increased by 1 unit, that is $t=3$,

$\begin{array}{c}{y}_2=A{e}^{0.1\times 3}\\ =A{e}^{0.3}\\ \approx 1.35A\end{array}$

Hence, the percentage increase of both values is approximately 10.57%.

Therefore, the given statement is false.

To determine

To explain: Whether the statement $\ln xy=\left(\ln x\right)\left(\ln y\right)$ is true or false.

Explanation of Solution

Consider the equation $\ln xy=\left(\ln x\right)\left(\ln y\right)$.

Suppose $x=2\text{ and y}=3$.

Compute the value of $\ln xy$.

$\begin{array}{c}\ln \left(2\times 3\right)=\ln 6\\ \approx 1.791\end{array}$

Compute the value of $\left(\ln x\right)\left(\ln y\right)$.

$\begin{array}{c}\left(\ln x\right)\left(\ln y\right)=\left(\ln 2\right)\left(\ln 3\right)\\ =0.693\times 1.09\\ \approx 0.762\end{array}$

Clearly, $\ln 6\ne \left(\ln 2\right)\left(\ln 3\right)$.

Therefore, the given statement is false.

To determine

To explain: Whether the statement $\sinh \left(\ln x\right)=\frac{x^2-1}{2x}$ is true or false.

Explanation of Solution

Compute $\sinh \left(\ln x\right)$ as follows.

$\begin{array}{c}\sinh \left(\ln x\right)=\frac{e^{\ln x}-e^{-\ln x}}{2}\left[\because \sinh x=\frac{e^x-e^{-x}}{2}\right]\\ =\frac{e^{\ln x}-e^{\ln x^{-1}}}{2}\left[\because \ln x^a=a\ln x\right]\\ =\frac{x-\frac{1}{x}}{2}\left[\because e^{\ln x}=x\right]\\ =\frac{x^2-1}{2x}\end{array}$

Hence, $\sinh \left(\ln x\right)=\frac{x^2-1}{2x}$.

Therefore, the given statement is true.