Algebra Examples

$y = {(2^{x} - 3)}^{\frac{1}{4}}$

Step 1

Interchange the variables.

$x = {(2^{y} - 3)}^{\frac{1}{4}}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as ${(2^{y} - 3)}^{\frac{1}{4}} = x$ .

${(2^{y} - 3)}^{\frac{1}{4}} = x$

Step 2.2

Raise each side of the equation to the power of $4$ to eliminate the fractional exponent on the left side.

${({(2^{y} - 3)}^{\frac{1}{4}})}^{4} = x^{4}$

Step 2.3

Simplify the left side.

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Step 2.3.1

Simplify ${({(2^{y} - 3)}^{\frac{1}{4}})}^{4}$ .

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Step 2.3.1.1

Multiply the exponents in ${({(2^{y} - 3)}^{\frac{1}{4}})}^{4}$ .

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Step 2.3.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(2^{y} - 3)}^{\frac{1}{4} \cdot 4} = x^{4}$

Step 2.3.1.1.2

Cancel the common factor of $4$ .

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Step 2.3.1.1.2.1

Cancel the common factor.

${(2^{y} - 3)}^{\frac{1}{4} \cdot 4} = x^{4}$

Step 2.3.1.1.2.2

Rewrite the expression.

${(2^{y} - 3)}^{1} = x^{4}$

Step 2.3.1.2

Simplify.

$2^{y} - 3 = x^{4}$

Step 2.4

Solve for $y$ .

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Step 2.4.1

Add $3$ to both sides of the equation.

$2^{y} = x^{4} + 3$

Step 2.4.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (2^{y}) = \ln (x^{4} + 3)$

Step 2.4.3

Expand $\ln (2^{y})$ by moving $y$ outside the logarithm.

$y \ln (2) = \ln (x^{4} + 3)$

Step 2.4.4

Divide each term in $y \ln (2) = \ln (x^{4} + 3)$ by $\ln (2)$ and simplify.

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Step 2.4.4.1

Divide each term in $y \ln (2) = \ln (x^{4} + 3)$ by $\ln (2)$ .

$\frac{y \ln (2)}{\ln (2)} = \frac{\ln (x^{4} + 3)}{\ln (2)}$

Step 2.4.4.2

Simplify the left side.

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Step 2.4.4.2.1

Cancel the common factor of $\ln (2)$ .

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Step 2.4.4.2.1.1

Cancel the common factor.

$\frac{y \ln (2)}{\ln (2)} = \frac{\ln (x^{4} + 3)}{\ln (2)}$

Step 2.4.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (x^{4} + 3)}{\ln (2)}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (x^{4} + 3)}{\ln (2)}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (x^{4} + 3)}{\ln (2)}$ is the inverse of $f (x) = {(2^{x} - 3)}^{\frac{1}{4}}$ .

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate $f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{\ln ({({(2^{x} - 3)}^{\frac{1}{4}})}^{4} + 3)}{\ln (2)}$

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

Simplify each term.

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Step 4.2.3.1.1

Multiply the exponents in ${({(2^{x} - 3)}^{\frac{1}{4}})}^{4}$ .

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Step 4.2.3.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{\ln ({(2^{x} - 3)}^{\frac{1}{4} \cdot 4} + 3)}{\ln (2)}$

Step 4.2.3.1.1.2

Cancel the common factor of $4$ .

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Step 4.2.3.1.1.2.1

Cancel the common factor.

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{\ln ({(2^{x} - 3)}^{\frac{1}{4} \cdot 4} + 3)}{\ln (2)}$

Step 4.2.3.1.1.2.2

Rewrite the expression.

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{\ln ((2^{x} - 3) + 3)}{\ln (2)}$

Step 4.2.3.1.2

Simplify.

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{\ln (2^{x} - 3 + 3)}{\ln (2)}$

Step 4.2.3.2

Combine the opposite terms in $2^{x} - 3 + 3$ .

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Step 4.2.3.2.1

Add $- 3$ and $3$ .

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{\ln (2^{x} + 0)}{\ln (2)}$

Step 4.2.3.2.2

Add $2^{x}$ and $0$ .

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{\ln (2^{x})}{\ln (2)}$

Step 4.2.4

Expand $\ln (2^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{x \ln (2)}{\ln (2)}$

Step 4.2.5

Cancel the common factor of $\ln (2)$ .

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Step 4.2.5.1

Cancel the common factor.

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = \frac{x \ln (2)}{\ln (2)}$

Step 4.2.5.2

Divide $x$ by $1$ .

$f^{- 1} ({(2^{x} - 3)}^{\frac{1}{4}}) = x$

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate $f (\frac{\ln (x^{4} + 3)}{\ln (2)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x^{4} + 3)}{\ln (2)}) = {(2^{\frac{\ln (x^{4} + 3)}{\ln (2)}} - 3)}^{\frac{1}{4}}$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Use the change of base rule $\frac{\log_{b} (x)}{\log_{b} (a)} = \log_{a} (x)$ .

$f (\frac{\ln (x^{4} + 3)}{\ln (2)}) = {(2^{\log_{2} (x^{4} + 3)} - 3)}^{\frac{1}{4}}$

Step 4.3.3.2

Exponentiation and log are inverse functions.

$f (\frac{\ln (x^{4} + 3)}{\ln (2)}) = {(x^{4} + 3 - 3)}^{\frac{1}{4}}$

Step 4.3.4

Simplify by adding terms.

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Step 4.3.4.1

Combine the opposite terms in $x^{4} + 3 - 3$ .

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Step 4.3.4.1.1

Subtract $3$ from $3$ .

$f (\frac{\ln (x^{4} + 3)}{\ln (2)}) = {(x^{4} + 0)}^{\frac{1}{4}}$

Step 4.3.4.1.2

Add $x^{4}$ and $0$ .

$f (\frac{\ln (x^{4} + 3)}{\ln (2)}) = {(x^{4})}^{\frac{1}{4}}$

Step 4.3.4.2

Multiply the exponents in ${(x^{4})}^{\frac{1}{4}}$ .

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Step 4.3.4.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{\ln (x^{4} + 3)}{\ln (2)}) = x^{4 (\frac{1}{4})}$

Step 4.3.4.2.2

Cancel the common factor of $4$ .

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Step 4.3.4.2.2.1

Cancel the common factor.

$f (\frac{\ln (x^{4} + 3)}{\ln (2)}) = x^{4 (\frac{1}{4})}$

Step 4.3.4.2.2.2

Rewrite the expression.

$f (\frac{\ln (x^{4} + 3)}{\ln (2)}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x^{4} + 3)}{\ln (2)}$ is the inverse of $f (x) = {(2^{x} - 3)}^{\frac{1}{4}}$ .

$f^{- 1} (x) = \frac{\ln (x^{4} + 3)}{\ln (2)}$