Pre-Algebra Examples

$5 \log_{7} (x - 2)$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

${(x - 2)}^{5} = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 1.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.3

The vertical asymptote occurs at $x = 2$ .

Vertical Asymptote: $x = 2$

Step 2

Find the point at $x = 3$ .

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Step 2.1

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

$f (3) = 5 \log_{7} ((3) - 2)$

Step 2.2

Simplify the result.

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Step 2.2.1

Subtract $2$ from $3$ .

$f (3) = 5 \log_{7} (1)$

Step 2.2.2

Logarithm base $7$ of $1$ is $0$ .

$f (3) = 5 \cdot 0$

Step 2.2.3

Multiply $5$ by $0$ .

$f (3) = 0$

Step 2.2.4

The final answer is $0$ .

$0$

Step 2.3

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = 9$ .

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Step 3.1

Replace the variable $x$ with $9$ in the expression.

$f (9) = 5 \log_{7} ((9) - 2)$

Step 3.2

Simplify the result.

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Step 3.2.1

Subtract $2$ from $9$ .

$f (9) = 5 \log_{7} (7)$

Step 3.2.2

Logarithm base $7$ of $7$ is $1$ .

$f (9) = 5 \cdot 1$

Step 3.2.3

Multiply $5$ by $1$ .

$f (9) = 5$

Step 3.2.4

The final answer is $5$ .

$5$

Step 3.3

Convert $5$ to decimal.

$y = 5$

Step 4

Find the point at $x = 4$ .

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Step 4.1

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

$f (4) = 5 \log_{7} ((4) - 2)$

Step 4.2

Simplify the result.

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Step 4.2.1

Subtract $2$ from $4$ .

$f (4) = 5 \log_{7} (2)$

Step 4.2.2

Simplify $5 \log_{7} (2)$ by moving $5$ inside the logarithm.

$f (4) = \log_{7} (2^{5})$

Step 4.2.3

Raise $2$ to the power of $5$ .

$f (4) = \log_{7} (32)$

Step 4.2.4

The final answer is $\log_{7} (32)$ .

$\log_{7} (32)$

Step 4.3

Convert $\log_{7} (32)$ to decimal.

$y = 1.78103593$

Step 5

The log function can be graphed using the vertical asymptote at $x = 2$ and the points $(3, 0), (9, 5), (4, 1.78103593)$ .

Vertical Asymptote: $x = 2$

$\begin{array}{c} x & y \\ 3 & 0 \\ 4 & 1.781 \\ 9 & 5 \end{array}$

Step 6