Graph log base 3 of x+1

$\log_{3} (x + 1)$

Step 1

Find the asymptotes.

Set the argument of the logarithm equal to zero.

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The vertical asymptote occurs at $x = - 1$ .

Vertical Asymptote: $x = - 1$

Step 2

Find the point at $x = 0$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = \log_{3} ((0) + 1)$

Simplify the result.

Add $0$ and $1$ .

$f (0) = \log_{3} (1)$

Logarithm base $3$ of $1$ is $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = 2$ .

Replace the variable $x$ with $2$ in the expression.

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

Simplify the result.

Add $2$ and $1$ .

$f (2) = \log_{3} (3)$

Logarithm base $3$ of $3$ is $1$ .

$f (2) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Step 4

Find the point at $x = 8$ .

Replace the variable $x$ with $8$ in the expression.

$f (8) = \log_{3} ((8) + 1)$

Simplify the result.

Add $8$ and $1$ .

$f (8) = \log_{3} (9)$

Logarithm base $3$ of $9$ is $2$ .

Rewrite as an equation.

$\log_{3} (9) = x$

Rewrite $\log_{3} (9) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$3^{x} = 9$

Create equivalent expressions in the equation that all have equal bases.

$3^{x} = 3^{2}$

Since the bases are the same, the two expressions are only equal if the exponents are also equal.

$x = 2$

The variable $x$ is equal to $2$ .

$f (8) = 2$

The final answer is $2$ .

$2$

Convert $2$ to decimal.

$y = 2$

Step 5

The log function can be graphed using the vertical asymptote at $x = - 1$ and the points $(0, 0), (2, 1), (8, 2)$ .

Vertical Asymptote: $x = - 1$

$\begin{array}{c} x & y \\ 0 & 0 \\ 2 & 1 \\ 8 & 2 \end{array}$

Step 6

Graph log base 3 of x+1

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