Graph log base 2 of x+1

$\log_{2} (x + 1)$

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.

$f (0) = \log_{2} ((0) + 1)$

Simplify the result.

Add $0$ and $1$ .

$f (0) = \log_{2} (1)$

Logarithm base $2$ 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 $x = 1$ .

Replace the variable $x$ with $1$ in the expression.

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

Simplify the result.

Add $1$ and $1$ .

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

Logarithm base $2$ of $2$ is $1$ .

$f (1) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Step 4

Find the point at $x = 3$ .

Replace the variable $x$ with $3$ in the expression.

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

Simplify the result.

Add $3$ and $1$ .

$f (3) = \log_{2} (4)$

Logarithm base $2$ of $4$ is $2$ .

Rewrite as an equation.

$\log_{2} (4) = x$

Rewrite $\log_{2} (4) = x$ 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 $b^{y} = x$ .

$2^{x} = 4$

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

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

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 $2$ .

$f (3) = 2$

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 $(0, 0), (1, 1), (3, 2)$ .

Vertical Asymptote: $x = - 1$

$\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 1 \\ 3 & 2 \end{array}$

Step 6

Graph log base 2 of x+1

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