Graph y = log base 5 of x+1

$y = \log_{5} (x + 1)$

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.

$f (0) = \log_{5} ((0) + 1)$

Simplify the result.

Add $0$ and $1$ .

$f (0) = \log_{5} (1)$

Logarithm base $5$ 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 $x = 4$ .

Replace the variable $x$ with $4$ in the expression.

$f (4) = \log_{5} ((4) + 1)$

Simplify the result.

Add $4$ and $1$ .

$f (4) = \log_{5} (5)$

Logarithm base $5$ of $5$ is $1$ .

$f (4) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Step 4

Find the point at $x = 1$ .

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

$f (1) = \log_{5} ((1) + 1)$

Simplify the result.

Add $1$ and $1$ .

$f (1) = \log_{5} (2)$

The final answer is $\log_{5} (2)$ .

$\log_{5} (2)$

Convert $\log_{5} (2)$ to decimal.

$y = 0.43067655$

Step 5

The log function can be graphed using the vertical asymptote at $x = - 1$ and the points $(0, 0), (4, 1), (1, 0.43067655)$ .

Vertical Asymptote: $x = - 1$

$\begin{array}{c} x & y \\ 0 & 0 \\ 1 & 0.431 \\ 4 & 1 \end{array}$

Step 6

Graph y = log base 5 of x+1

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