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

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

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

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

Add $2$ to both sides of the equation.

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

Rewrite $\log_{\frac{1}{3}} (x + 1) = 2$ 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 $b^{y} = x$ .

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

Solve for $x$ .

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

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

Simplify ${(\frac{1}{3})}^{2}$ .

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

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

One to any power is one.

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

Raise $3$ to the power of $2$ .

$x + 1 = \frac{1}{9}$

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

Subtract $1$ from both sides of the equation.

$x = \frac{1}{9} - 1$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$x = \frac{1}{9} - 1 \cdot \frac{9}{9}$

Combine $- 1$ and $\frac{9}{9}$ .

$x = \frac{1}{9} + \frac{- 1 \cdot 9}{9}$

Combine the numerators over the common denominator.

$x = \frac{1 - 1 \cdot 9}{9}$

Simplify the numerator.

Multiply $- 1$ by $9$ .

$x = \frac{1 - 9}{9}$

Subtract $9$ from $1$ .

$x = \frac{- 8}{9}$

Move the negative in front of the fraction.

$x = - \frac{8}{9}$

x-intercept(s) in point form.

x-intercept(s): $(- \frac{8}{9}, 0)$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

Add $0$ and $1$ .

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

Logarithm base $\frac{1}{3}$ of $1$ is $0$ .

$y = 0 - 2$

Subtract $2$ from $0$ .

$y = - 2$

y-intercept(s) in point form.

y-intercept(s): $(0, - 2)$

Step 3

List the intersections.

x-intercept(s): $(- \frac{8}{9}, 0)$

y-intercept(s): $(0, - 2)$

Step 4

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

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