Find the Inverse y = log base 12 of 1/x

$y = \log_{12} (\frac{1}{x})$

Interchange the variables.

$x = \log_{12} (\frac{1}{y})$

Solve for $y$ .

Rewrite the equation as $\log_{12} (\frac{1}{y}) = x$ .

$\log_{12} (\frac{1}{y}) = x$

Rewrite $\log_{12} (\frac{1}{y}) = x$ 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 $b^{y} = x$ .

$12^{x} = \frac{1}{y}$

Solve for $y$ .

Rewrite the equation as $\frac{1}{y} = 12^{x}$ .

$\frac{1}{y} = 12^{x}$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y, 1$

The LCM of one and any expression is the expression.

$y$

Multiply each term in $\frac{1}{y} = 12^{x}$ by $y$ to eliminate the fractions.

Multiply each term in $\frac{1}{y} = 12^{x}$ by $y$ .

$\frac{1}{y} y = 12^{x} y$

Simplify the left side.

Cancel the common factor of $y$ .

Cancel the common factor.

$\frac{1}{y} y = 12^{x} y$

Rewrite the expression.

$1 = 12^{x} y$

Simplify the right side.

Reorder factors in $12^{x} y$ .

$1 = y \cdot 12^{x}$

Solve the equation.

Rewrite the equation as $y \cdot 12^{x} = 1$ .

$y \cdot 12^{x} = 1$

Divide each term in $(y \cdot 12^{x}) = 1$ by $12^{x}$ and simplify.

Divide each term in $(y \cdot 12^{x}) = 1$ by $12^{x}$ .

$\frac{(y \cdot 12^{x})}{12^{x}} = \frac{1}{12^{x}}$

Simplify the left side.

Cancel the common factor of $12^{x}$ .

Cancel the common factor.

$\frac{y \cdot 12^{x}}{12^{x}} = \frac{1}{12^{x}}$

Divide $y$ by $1$ .

$y = \frac{1}{12^{x}}$

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

$f^{- 1} (x) = \frac{1}{12^{x}}$

Verify if $f^{- 1} (x) = \frac{1}{12^{x}}$ is the inverse of $f (x) = \log_{12} (\frac{1}{x})$ .

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

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

Set up the composite result function.

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

Evaluate $f^{- 1} (\log_{12} (\frac{1}{x}))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\log_{12} (\frac{1}{x})) = \frac{1}{12^{\log_{12} (\frac{1}{x})}}$

Exponentiation and log are inverse functions.

$f^{- 1} (\log_{12} (\frac{1}{x})) = \frac{1}{\frac{1}{x}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\log_{12} (\frac{1}{x})) = 1 x$

Multiply $x$ by $1$ .

$f^{- 1} (\log_{12} (\frac{1}{x})) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{1}{12^{x}})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{1}{12^{x}}) = \log_{12} (\frac{1}{\frac{1}{12^{x}}})$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{1}{12^{x}}) = \log_{12} (1 \cdot 12^{x})$

Rewrite $\log_{12} (1 \cdot 12^{x})$ as $\log_{12} (1) + \log_{12} (12^{x})$ .

$f (\frac{1}{12^{x}}) = \log_{12} (1) + \log_{12} (12^{x})$

Use logarithm rules to move $x$ out of the exponent.

$f (\frac{1}{12^{x}}) = \log_{12} (1) + x \log_{12} (12)$

Logarithm base $12$ of $12$ is $1$ .

$f (\frac{1}{12^{x}}) = \log_{12} (1) + x \cdot 1$

Multiply $x$ by $1$ .

$f (\frac{1}{12^{x}}) = \log_{12} (1) + x$

Logarithm base $12$ of $1$ is $0$ .

$f (\frac{1}{12^{x}}) = 0 + x$

Add $0$ and $x$ .

$f (\frac{1}{12^{x}}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{1}{12^{x}}$ is the inverse of $f (x) = \log_{12} (\frac{1}{x})$ .

$f^{- 1} (x) = \frac{1}{12^{x}}$

Find the Inverse y = log base 12 of 1/x

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