Graph f(x) = log base 5 of 4x+1

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Trigonometry

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

Find the asymptotes.

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Set the argument of the logarithm equal to zero.

$4 x + 1 = 0$

Solve for $x$ .

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Subtract $1$ from both sides of the equation.

$4 x = - 1$

Divide each term by $4$ and simplify.

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Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

The vertical asymptote occurs at $x = - \frac{1}{4}$ .

Vertical Asymptote: $x = - \frac{1}{4}$

Find the point at $x = 1$ .

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

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

Simplify the result.

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Multiply $4$ by $1$ .

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

Add $4$ and $1$ .

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

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

$f (1) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Find the point at $x = 6$ .

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Replace the variable $x$ with $6$ in the expression.

$f (6) = \log_{5} (4 (6) + 1)$

Simplify the result.

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Multiply $4$ by $6$ .

$f (6) = \log_{5} (24 + 1)$

Add $24$ and $1$ .

$f (6) = \log_{5} (25)$

Logarithm base $5$ of $25$ is $2$ .

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Rewrite as an equation.

$\log_{5} (25) = x$

Rewrite $\log_{5} (25) = x$ 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 $b^{y} = x$ .

$5^{x} = 25$

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

$5^{x} = 5^{2}$

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 $2$ .

$f (6) = 2$

The final answer is $2$ .

$2$

Convert $2$ to decimal.

$y = 2$

Find the point at $x = 2$ .

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

$f (2) = \log_{5} (4 (2) + 1)$

Simplify the result.

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Multiply $4$ by $2$ .

$f (2) = \log_{5} (8 + 1)$

Add $8$ and $1$ .

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

The final answer is $\log_{5} (9)$ .

$\log_{5} (9)$

Convert $\log_{5} (9)$ to decimal.

$y = 1.36521238$

The log function can be graphed using the vertical asymptote at $x = - \frac{1}{4}$ and the points $(1, 1), (6, 2), (2, 1.36521238)$ .

Vertical Asymptote: $x = - \frac{1}{4}$

$\begin{array}{c} x & y \\ 1 & 1 \\ 2 & 1.365 \\ 6 & 2 \end{array}$

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