91Ó°ÊÓ

$$\begin{gathered} Iff\left(x\right)={\log }_3(x+2)thenf\left(x\right)=0: \\ ⇒{\log }_3(x+2)=0 \end{gathered}$$

Logarithmic function is good defined only if the argument of logarithmic function is a positive real number (log_bx, x >0).

$For{\log }_3(x+2):x+2>0⇒x>-2$

Therefore, the solutions of this equation must be greater than$-2---\left(1\right)$

Using a formula:

$y={\log }_{a}x⇔x=a^{y}a>0,a≠1$

(where a = 3 and y = 0) we get

$$\begin{gathered} {\log }_3(x+2)=0 \\ x+2=3^0 \\ x=-1 \end{gathered}$$

The solution is greater than - 2 (because of (1)), so the solution of the given equation is$x=-1$

$$\begin{gathered} iff\left(x\right)={\log }_3(x+2)thenf\left(x\right)>0: \\ ⇒{\log }_3(x+2)>0 \end{gathered}$$

Logarithmic function is good defined only if the argument of logarithmic function is a positive real number (log_bx, x >0).

$For{\log }_3(x+2):x+2>0⇒x>-2$

Therefore, the solutions of this equation must be greater than role="math" localid="1647669047575" $-2---(*)$

In order to solve the inequality, it is necessary to write the expressions on the left and on the right using a logarithmic with the same base.

Therefore, we will write everything using the logarithm with base 3.

Using the property: $r={\log }_a{a}^r$we get the following $0={\log }_3{3}^0={\log }_{3}1$

So,

$$\begin{gathered} {\log }_3(x+2)>0 \\ {\log }_3(x+2)>{\log }_{3}1 \end{gathered}$$

For increasing functions, the following holds:

$Iff\left(x\right)>f\left(y\right)thenx>y---\left(1\right)$

Since the base a = 3 is greater than 1 then the logarithmic function is increasing $---\left(2\right)$

Using property (1) and (2) we conclude:

$$\begin{gathered} {\log }_3(x+2)>{\log }_{3}1 \\ x+2>1 \\ x>1-2 \\ x>-1 \end{gathered}$$

The solution to the given inequality is$x∈(-1,∞)=\left\{x\right|x>-1\}$

$$\begin{gathered} Iff\left(x\right)={\log }_3(x+2)thenf\left(x\right)=3: \\ ⇒{\log }_3(x+2)=3 \end{gathered}$$

Logarithmic function is good defined only if the argument of logarithmic function is a positive real number (log_bx, x >0).

$For{\log }_3(x+2):x+2>0⇒x>-2$

Therefore, the solutions of this equation must be greater thanrole="math" localid="1647669299314" $-2---(*)$

Using a formula:

$y={\log }_{a}x⇔x=a^{y}a>0,a≠1$

(where a = 3 and y = 3) we get:

$$\begin{gathered} {\log }_3(x+2)=3 \\ x+2=3^3 \\ x+2=27 \\ x=27-2 \\ x=25 \end{gathered}$$

The solution is greater than - 2 (because of (*)), so the solution of the given equation is x = 25.