Graph f(x)=3 log base 2 of x

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

Find the asymptotes.

Set the argument of the logarithm equal to zero.

$x^{3} = 0$

Solve for $x$ .

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

The vertical asymptote occurs at $x = 0$ .

Vertical Asymptote: $x = 0$

Find the point at $x = 1$ .

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

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

Simplify the result.

Logarithm base $2$ of $1$ is $0$ .

$f (1) = 3 \cdot 0$

Multiply $3$ by $0$ .

$f (1) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Find the point at $x = 2$ .

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

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

Simplify the result.

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

$f (2) = 3 \cdot 1$

Multiply $3$ by $1$ .

$f (2) = 3$

The final answer is $3$ .

$3$

Convert $3$ to decimal.

$y = 3$

Find the point at $x = 4$ .

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

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

Simplify the result.

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

$f (4) = 3 \cdot 2$

Multiply $3$ by $2$ .

$f (4) = 6$

The final answer is $6$ .

$6$

Convert $6$ to decimal.

$y = 6$

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 0), (2, 3), (4, 6)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 3 \\ 4 & 6 \end{array}$

Graph f(x)=3 log base 2 of x

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