Algebra Examples

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

Step 1

Find the x-intercepts.

Tap for more steps...

Step 1.1

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

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

Step 1.2

Solve the equation.

Tap for more steps...

Step 1.2.1

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

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

Step 1.2.2

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

Tap for more steps...

Step 1.2.2.1

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

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

Step 1.2.2.2

Simplify the left side.

Tap for more steps...

Step 1.2.2.2.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 1.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.2.2.1.2

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

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

Step 1.2.2.3

Simplify the right side.

Tap for more steps...

Step 1.2.2.3.1

Divide $0$ by $4$ .

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

Step 1.2.3

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 $b^{y} = x$ .

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

Step 1.2.4

Solve for $x$ .

Tap for more steps...

Step 1.2.4.1

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

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

Step 1.2.4.2

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

Tap for more steps...

Step 1.2.4.2.1

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

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

Step 1.2.4.2.2

Anything raised to $0$ is $1$ .

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

Step 1.2.4.2.3

Anything raised to $0$ is $1$ .

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

Step 1.2.4.2.4

Divide $1$ by $1$ .

$x + 2 = 1$

Step 1.2.4.3

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

Tap for more steps...

Step 1.2.4.3.1

Subtract $2$ from both sides of the equation.

$x = 1 - 2$

Step 1.2.4.3.2

Subtract $2$ from $1$ .

$x = - 1$

Step 1.3

x-intercept(s) in point form.

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

Step 2

Find the y-intercepts.

Tap for more steps...

Step 2.1

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

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

Step 2.2

Solve the equation.

Tap for more steps...

Step 2.2.1

Remove parentheses.

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

Step 2.2.2

Remove parentheses.

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

Step 2.2.3

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

Tap for more steps...

Step 2.2.3.1

Add $0$ and $2$ .

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

Step 2.2.3.2

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

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

Step 2.2.3.3

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

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

Step 2.3

y-intercept(s) in point form.

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

Step 3

List the intersections.

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

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

Step 4