Find the Inverse 10^(x-1)

$10^{x - 1}$

Step 1

Interchange the variables.

$x = 10^{y - 1}$

Step 2

Solve for $y$ .

Rewrite the equation as $10^{y - 1} = x$ .

$10^{y - 1} = x$

Take the base $10$ logarithm of both sides of the equation to remove the variable from the exponent.

$\log (10^{y - 1}) = \log (x)$

Expand the left side.

Expand $\log (10^{y - 1})$ by moving $y - 1$ outside the logarithm.

$(y - 1) \log (10) = \log (x)$

Logarithm base $10$ of $10$ is $1$ .

$(y - 1) \cdot 1 = \log (x)$

Multiply $y - 1$ by $1$ .

$y - 1 = \log (x)$

Add $1$ to both sides of the equation.

$y = \log (x) + 1$

Step 3

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

$f^{- 1} (x) = \log (x) + 1$

Step 4

Verify if $f^{- 1} (x) = \log (x) + 1$ is the inverse of $f (x) = 10^{x - 1}$ .

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 $f^{- 1} \cdot 10^{x - 1}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot 10^{x - 1} = \log (10^{x - 1}) + 1$

Simplify each term.

Use logarithm rules to move $x - 1$ out of the exponent.

$f^{- 1} \cdot 10^{x - 1} = (x - 1) \log (10) + 1$

Logarithm base $10$ of $10$ is $1$ .

$f^{- 1} \cdot 10^{x - 1} = (x - 1) \cdot 1 + 1$

Multiply $x - 1$ by $1$ .

$f^{- 1} \cdot 10^{x - 1} = x - 1 + 1$

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

Add $- 1$ and $1$ .

$f^{- 1} \cdot 10^{x - 1} = x + 0$

Add $x$ and $0$ .

$f^{- 1} \cdot 10^{x - 1} = x$

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

Set up the composite result function.

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

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

$f (\log (x) + 1) = 10^{(\log (x) + 1) - 1}$

Combine the opposite terms in $(\log (x) + 1) - 1$ .

Subtract $1$ from $1$ .

$f (\log (x) + 1) = 10^{\log (x) + 0}$

Add $\log (x)$ and $0$ .

$f (\log (x) + 1) = 10^{\log (x)}$

Exponentiation and log are inverse functions.

$f (\log (x) + 1) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \log (x) + 1$ is the inverse of $f (x) = 10^{x - 1}$ .

$f^{- 1} (x) = \log (x) + 1$

Find the Inverse 10^(x-1)

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