Chapter 5, Problem 15CR

(a)

To determine

To calculate: The solution of $f\left(x\right)=0$ for $f\left(x\right)={\log }_3\left(x+2\right)$ .

Answer to Problem 15CR

Solution:

The solution of $f\left(x\right)=0$ for $f\left(x\right)={\log }_3\left(x+2\right)$ is $\underset{\_}{x=-1}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)={\log }_3\left(x+2\right)$ .

Formula used:

Logarithm to exponential rule: if ${\log }_a\left(b\right)=c$ then $b=a^c$ .

Calculation:

To find the solution of $f\left(x\right)=0$ , set $f\left(x\right)={\log }_3\left(x+2\right)$ equals to 0.

  $\Rightarrow {\log }_3\left(x+2\right)=0$

Apply logarithm to exponential rule,

  $\Rightarrow x+2=3^0$

  $\Rightarrow x+2=1$

Subtract $2$ from both sides,

  $\Rightarrow x+2-2=1-2$

  $\Rightarrow x=-1$

Therefore, $x=-1$ is the solution for $f\left(x\right)=0$ .

(b)

To determine

The solution of an inequality $f\left(x\right)>0$ for $f\left(x\right)={\log }_3\left(x+2\right)$ .

Answer to Problem 15CR

Solution:

The solution of $f\left(x\right)>0$ for $f\left(x\right)={\log }_3\left(x+2\right)$ is $\underset{\_}{\left(-1,\infty \right)}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)={\log }_3\left(x+2\right)$ .

First, find real zeroes of equation $f\left(x\right)=0$ .

From part (a), the real zero of ${\log }_3\left(x+2\right)=0$ is $x=-1$ .

The logarithmic function is defined for positive values.

Therefore, the domain of $f\left(x\right)={\log }_3\left(x+2\right)$ is $x+2>0\Rightarrow x>-2$ .

Therefore, the intervals using number $x=-1$ are $\left(-2,-1\right),\left(-1,\infty \right)$ .

  $\begin{array}{ccc}\text{Intervals}& \left(-2,-1\right)& \left(-1,\infty \right)\\ \text{Test points}& -\frac{3}{2}& 2\\ \text{Value of }f\left(x\right)& -\text{0}\text{.63}& \text{1}\text{.26}\\ \text{Sign of }f\left(x\right)& \text{Negative}& \text{Positive}\end{array}$

Therefore, the solution of $f\left(x\right)>0$ is $\left(-1,\infty \right)$ .

(c)

To determine

To calculate: The solution of equation $f\left(x\right)=3$ for $f\left(x\right)={\log }_3\left(x+2\right)$ .

Answer to Problem 15CR

Solution:

The solution of $f\left(x\right)=3$ for $f\left(x\right)={\log }_3\left(x+2\right)$ is $\underset{\_}{x=25}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)={\log }_3\left(x+2\right)$ .

Formula used:

Logarithm to exponential rule: if ${\log }_a\left(b\right)=c$ then $b=a^c$ .

Calculation:

Set equation $f\left(x\right)={\log }_3\left(x+2\right)$ equals to $3$ ,

  $\Rightarrow {\log }_3\left(x+2\right)=3$

Apply logarithm to exponential rule,

  $\Rightarrow x+2=3^3$

  $\Rightarrow x+2=27$

Subtract $2$ from both sides,

  $\Rightarrow x+2-2=27-2$

  $\Rightarrow x=25$

Therefore, $x=25$ is the solution for $f\left(x\right)=3$ .