Precalculus Examples

$f (x) = \log_{3} (x + 5)$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$x + 5 = 0$

Step 1.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 1.3

The vertical asymptote occurs at $x = - 5$ .

Vertical Asymptote: $x = - 5$

Step 2

Find the point at $x = - 4$ .

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Step 2.1

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

$f (- 4) = \log_{3} ((- 4) + 5)$

Step 2.2

Simplify the result.

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Step 2.2.1

Add $- 4$ and $5$ .

$f (- 4) = \log_{3} (1)$

Step 2.2.2

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

$f (- 4) = 0$

Step 2.2.3

The final answer is $0$ .

$0$

Step 2.3

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = - 2$ .

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Step 3.1

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

$f (- 2) = \log_{3} ((- 2) + 5)$

Step 3.2

Simplify the result.

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Step 3.2.1

Add $- 2$ and $5$ .

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

Step 3.2.2

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

$f (- 2) = 1$

Step 3.2.3

The final answer is $1$ .

$1$

Step 3.3

Convert $1$ to decimal.

$y = 1$

Step 4

Find the point at $x = 4$ .

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Step 4.1

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

$f (4) = \log_{3} ((4) + 5)$

Step 4.2

Simplify the result.

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Step 4.2.1

Add $4$ and $5$ .

$f (4) = \log_{3} (9)$

Step 4.2.2

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

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Step 4.2.2.1

Rewrite as an equation.

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

Step 4.2.2.2

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 $b^{y} = x$ .

$3^{x} = 9$

Step 4.2.2.3

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

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

Step 4.2.2.4

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

$x = 2$

Step 4.2.2.5

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

$f (4) = 2$

Step 4.2.3

The final answer is $2$ .

$2$

Step 4.3

Convert $2$ to decimal.

$y = 2$

Step 5

The log function can be graphed using the vertical asymptote at $x = - 5$ and the points $(- 4, 0), (- 2, 1), (4, 2)$ .

Vertical Asymptote: $x = - 5$

$\begin{array}{c} x & y \\ - 4 & 0 \\ - 2 & 1 \\ 4 & 2 \end{array}$

Step 6