Find the Inverse y = log base 5 of 2x

$y = \log_{5} (2 x)$

Step 1

Interchange the variables.

$x = \log_{5} (2 y)$

Step 2

Solve for $y$ .

Rewrite the equation as $\log_{5} (2 y) = x$ .

$\log_{5} (2 y) = x$

Rewrite $\log_{5} (2 y) = x$ 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 $b^{y} = x$ .

$5^{x} = 2 y$

Solve for $y$ .

Rewrite the equation as $2 y = 5^{x}$ .

$2 y = 5^{x}$

Divide each term in $2 y = 5^{x}$ by $2$ and simplify.

Divide each term in $2 y = 5^{x}$ by $2$ .

$\frac{2 y}{2} = \frac{5^{x}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{5^{x}}{2}$

Divide $y$ by $1$ .

$y = \frac{5^{x}}{2}$

Step 3

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

$f^{- 1} (x) = \frac{5^{x}}{2}$

Step 4

Verify if $f^{- 1} (x) = \frac{5^{x}}{2}$ is the inverse of $f (x) = \log_{5} (2 x)$ .

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 $f^{- 1} (\log_{5} (2 x))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\log_{5} (2 x)) = \frac{5^{\log_{5} (2 x)}}{2}$

Exponentiation and log are inverse functions.

$f^{- 1} (\log_{5} (2 x)) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\log_{5} (2 x)) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$f^{- 1} (\log_{5} (2 x)) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{5^{x}}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5^{x}}{2}) = \log_{5} (2 (\frac{5^{x}}{2}))$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{5^{x}}{2}) = \log_{5} (2 (\frac{5^{x}}{2}))$

Rewrite the expression.

$f (\frac{5^{x}}{2}) = \log_{5} (5^{x})$

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

$f (\frac{5^{x}}{2}) = x \log_{5} (5)$

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

$f (\frac{5^{x}}{2}) = x \cdot 1$

Multiply $x$ by $1$ .

$f (\frac{5^{x}}{2}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5^{x}}{2}$ is the inverse of $f (x) = \log_{5} (2 x)$ .

$f^{- 1} (x) = \frac{5^{x}}{2}$

Find the Inverse y = log base 5 of 2x

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