Trigonometry Examples

$h (x) = 4 \log_{5} (x - 6) + 1$

Step 1

Write $h (x) = 4 \log_{5} (x - 6) + 1$ as an equation.

$y = 4 \log_{5} (x - 6) + 1$

Step 2

Interchange the variables.

$x = 4 \log_{5} (y - 6) + 1$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $4 \log_{5} (y - 6) + 1 = x$ .

$4 \log_{5} (y - 6) + 1 = x$

Step 3.2

Subtract $1$ from both sides of the equation.

$4 \log_{5} (y - 6) = x - 1$

Step 3.3

Divide each term in $4 \log_{5} (y - 6) = x - 1$ by $4$ and simplify.

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Step 3.3.1

Divide each term in $4 \log_{5} (y - 6) = x - 1$ by $4$ .

$\frac{4 \log_{5} (y - 6)}{4} = \frac{x}{4} + \frac{- 1}{4}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $4$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{4 \log_{5} (y - 6)}{4} = \frac{x}{4} + \frac{- 1}{4}$

Step 3.3.2.1.2

Divide $\log_{5} (y - 6)$ by $1$ .

$\log_{5} (y - 6) = \frac{x}{4} + \frac{- 1}{4}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Move the negative in front of the fraction.

$\log_{5} (y - 6) = \frac{x}{4} - \frac{1}{4}$

Step 3.4

Rewrite $\log_{5} (y - 6) = \frac{x}{4} - \frac{1}{4}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$5^{\frac{x}{4} - \frac{1}{4}} = y - 6$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $y - 6 = 5^{\frac{x}{4} - \frac{1}{4}}$ .

$y - 6 = 5^{\frac{x}{4} - \frac{1}{4}}$

Step 3.5.2

Add $6$ to both sides of the equation.

$y = 5^{\frac{x}{4} - \frac{1}{4}} + 6$

Step 4

Replace $y$ with $h^{- 1} (x)$ to show the final answer.

$h^{- 1} (x) = 5^{\frac{x}{4} - \frac{1}{4}} + 6$

Step 5

Verify if $h^{- 1} (x) = 5^{\frac{x}{4} - \frac{1}{4}} + 6$ is the inverse of $h (x) = 4 \log_{5} (x - 6) + 1$ .

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Step 5.1

To verify the inverse, check if $h^{- 1} (h (x)) = x$ and $h (h^{- 1} (x)) = x$ .

Step 5.2

Evaluate $h^{- 1} (h (x))$ .

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Step 5.2.1

Set up the composite result function.

$h^{- 1} (h (x))$

Step 5.2.2

Evaluate $h^{- 1} (4 \log_{5} (x - 6) + 1)$ by substituting in the value of $h$ into $h^{- 1}$ .

$h^{- 1} (4 \log_{5} (x - 6) + 1) = 5^{\frac{4 \log_{5} (x - 6) + 1}{4} - \frac{1}{4}} + 6$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Combine the numerators over the common denominator.

$h^{- 1} (4 \log_{5} (x - 6) + 1) = 5^{\frac{4 \log_{5} (x - 6) + 1 - 1}{4}} + 6$

Step 5.2.3.2

Combine the opposite terms in $4 \log_{5} (x - 6) + 1 - 1$ .

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Step 5.2.3.2.1

Subtract $1$ from $1$ .

$h^{- 1} (4 \log_{5} (x - 6) + 1) = 5^{\frac{4 \log_{5} (x - 6) + 0}{4}} + 6$

Step 5.2.3.2.2

Add $4 \log_{5} (x - 6)$ and $0$ .

$h^{- 1} (4 \log_{5} (x - 6) + 1) = 5^{\frac{4 \log_{5} (x - 6)}{4}} + 6$

Step 5.2.3.3

Simplify $4 \log_{5} (x - 6)$ by moving $4$ inside the logarithm.

$h^{- 1} (4 \log_{5} (x - 6) + 1) = 5^{\frac{\log_{5} ({(x - 6)}^{4})}{4}} + 6$

Step 5.2.3.4

Expand $\log_{5} ({(x - 6)}^{4})$ by moving $4$ outside the logarithm.

$h^{- 1} (4 \log_{5} (x - 6) + 1) = 5^{\frac{4 \log_{5} (x - 6)}{4}} + 6$

Step 5.2.3.5

Cancel the common factor of $4$ .

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Step 5.2.3.5.1

Cancel the common factor.

$h^{- 1} (4 \log_{5} (x - 6) + 1) = 5^{\frac{4 \log_{5} (x - 6)}{4}} + 6$

Step 5.2.3.5.2

Divide $\log_{5} (x - 6)$ by $1$ .

$h^{- 1} (4 \log_{5} (x - 6) + 1) = 5^{\log_{5} (x - 6)} + 6$

Step 5.2.3.6

Exponentiation and log are inverse functions.

$h^{- 1} (4 \log_{5} (x - 6) + 1) = x - 6 + 6$

Step 5.2.4

Combine the opposite terms in $x - 6 + 6$ .

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Step 5.2.4.1

Add $- 6$ and $6$ .

$h^{- 1} (4 \log_{5} (x - 6) + 1) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$h^{- 1} (4 \log_{5} (x - 6) + 1) = x$

Step 5.3

Evaluate $h (h^{- 1} (x))$ .

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Step 5.3.1

Set up the composite result function.

$h (h^{- 1} (x))$

Step 5.3.2

Evaluate $h (5^{\frac{x}{4} - \frac{1}{4}} + 6)$ by substituting in the value of $h^{- 1}$ into $h$ .

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = 4 \log_{5} ((5^{\frac{x}{4} - \frac{1}{4}} + 6) - 6) + 1$

Step 5.3.3

Combine the opposite terms in $4 \log_{5} ((5^{\frac{x}{4} - \frac{1}{4}} + 6) - 6) + 1$ .

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Step 5.3.3.1

Subtract $6$ from $6$ .

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = 4 \log_{5} (5^{\frac{x}{4} - \frac{1}{4}} + 0) + 1$

Step 5.3.3.2

Add $5^{\frac{x}{4} - \frac{1}{4}}$ and $0$ .

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = 4 \log_{5} (5^{\frac{x}{4} - \frac{1}{4}}) + 1$

Step 5.3.4

Simplify each term.

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Step 5.3.4.1

Use logarithm rules to move $\frac{x}{4} - \frac{1}{4}$ out of the exponent.

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = 4 ((\frac{x}{4} - \frac{1}{4}) \log_{5} (5)) + 1$

Step 5.3.4.2

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

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = 4 ((\frac{x}{4} - \frac{1}{4}) \cdot 1) + 1$

Step 5.3.4.3

Multiply $\frac{x}{4} - \frac{1}{4}$ by $1$ .

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = 4 (\frac{x}{4} - \frac{1}{4}) + 1$

Step 5.3.4.4

Apply the distributive property.

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = 4 (\frac{x}{4}) + 4 (- \frac{1}{4}) + 1$

Step 5.3.4.5

Cancel the common factor of $4$ .

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Step 5.3.4.5.1

Cancel the common factor.

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = 4 (\frac{x}{4}) + 4 (- \frac{1}{4}) + 1$

Step 5.3.4.5.2

Rewrite the expression.

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = x + 4 (- \frac{1}{4}) + 1$

Step 5.3.4.6

Cancel the common factor of $4$ .

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Step 5.3.4.6.1

Move the leading negative in $- \frac{1}{4}$ into the numerator.

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = x + 4 (\frac{- 1}{4}) + 1$

Step 5.3.4.6.2

Cancel the common factor.

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = x + 4 (\frac{- 1}{4}) + 1$

Step 5.3.4.6.3

Rewrite the expression.

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = x - 1 + 1$

Step 5.3.5

Combine the opposite terms in $x - 1 + 1$ .

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Step 5.3.5.1

Add $- 1$ and $1$ .

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$h (5^{\frac{x}{4} - \frac{1}{4}} + 6) = x$

Step 5.4

Since $h^{- 1} (h (x)) = x$ and $h (h^{- 1} (x)) = x$ , then $h^{- 1} (x) = 5^{\frac{x}{4} - \frac{1}{4}} + 6$ is the inverse of $h (x) = 4 \log_{5} (x - 6) + 1$ .

$h^{- 1} (x) = 5^{\frac{x}{4} - \frac{1}{4}} + 6$