Find the x and y Intercepts f(x)=4 log base 1/3 of x+2

$f (x) = 4 \log_{\frac{1}{3}} (x + 2)$

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = 4 \log_{\frac{1}{3}} (x + 2)$

Solve the equation.

Rewrite the equation as $4 \log_{\frac{1}{3}} (x + 2) = 0$ .

$4 \log_{\frac{1}{3}} (x + 2) = 0$

Divide each term in $4 \log_{\frac{1}{3}} (x + 2) = 0$ by $4$ and simplify.

Divide each term in $4 \log_{\frac{1}{3}} (x + 2) = 0$ by $4$ .

$\frac{4 \log_{\frac{1}{3}} (x + 2)}{4} = \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 \log_{\frac{1}{3}} (x + 2)}{4} = \frac{0}{4}$

Divide $\log_{\frac{1}{3}} (x + 2)$ by $1$ .

$\log_{\frac{1}{3}} (x + 2) = \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

$\log_{\frac{1}{3}} (x + 2) = 0$

Rewrite $\log_{\frac{1}{3}} (x + 2) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

${(\frac{1}{3})}^{0} = x + 2$

Solve for $x$ .

Rewrite the equation as $x + 2 = {(\frac{1}{3})}^{0}$ .

$x + 2 = {(\frac{1}{3})}^{0}$

Simplify ${(\frac{1}{3})}^{0}$ .

Apply the product rule to $\frac{1}{3}$ .

$x + 2 = \frac{1^{0}}{3^{0}}$

Anything raised to $0$ is $1$ .

$x + 2 = \frac{1}{3^{0}}$

Anything raised to $0$ is $1$ .

$x + 2 = \frac{1}{1}$

Divide $1$ by $1$ .

$x + 2 = 1$

Move all terms not containing $x$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$x = 1 - 2$

Subtract $2$ from $1$ .

$x = - 1$

x-intercept(s) in point form.

x-intercept(s): $(- 1, 0)$

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = 4 \log_{\frac{1}{3}} ((0) + 2)$

Solve the equation.

Remove parentheses.

$y = 4 \log_{\frac{1}{3}} (0 + 2)$

Remove parentheses.

$y = 4 \log_{\frac{1}{3}} ((0) + 2)$

Simplify $4 \log_{\frac{1}{3}} ((0) + 2)$ .

Add $0$ and $2$ .

$y = 4 \log_{\frac{1}{3}} (2)$

Simplify $4 \log_{\frac{1}{3}} (2)$ by moving $4$ inside the logarithm.

$y = \log_{\frac{1}{3}} (2^{4})$

Raise $2$ to the power of $4$ .

$y = \log_{\frac{1}{3}} (16)$

y-intercept(s) in point form.

y-intercept(s): $(0, \log_{\frac{1}{3}} (16))$

List the intersections.

x-intercept(s): $(- 1, 0)$

y-intercept(s): $(0, \log_{\frac{1}{3}} (16))$

Find the x and y Intercepts f(x)=4 log base 1/3 of x+2

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