Find the Inverse g(x)=10^(x-1)

$g (x) = 10^{x - 1}$

Step 1

Write $g (x) = 10^{x - 1}$ as an equation.

$y = 10^{x - 1}$

Step 2

Interchange the variables.

$x = 10^{y - 1}$

Step 3

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 4

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

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

Step 5

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

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

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

Set up the composite result function.

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

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

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

Simplify each term.

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

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

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

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

Multiply $x - 1$ by $1$ .

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

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

Subtract $1$ from $1$ .

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

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

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

Exponentiation and log are inverse functions.

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

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

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

Find the Inverse g(x)=10^(x-1)

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