Algebra Examples

N = \sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}}

$N = \sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}}$

Step 1

Rewrite the equation as

\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}} = N

$\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}} = N$ .

\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}} = N

$\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}} = N$

Step 2

To remove the radical on the left side of the equation, raise both sides of the equation to the power of

b

$b$ .

{\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}}}^{b} = N^{b}

${\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}}}^{b} = N^{b}$

Step 3

Simplify each side of the equation.

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Step 3.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}}

$\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}}$ as

{(\frac{1 + 2^{b}}{1 + 2^{- b}})}^{\frac{1}{b}}

${(\frac{1 + 2^{b}}{1 + 2^{- b}})}^{\frac{1}{b}}$ .

{({(\frac{1 + 2^{b}}{1 + 2^{- b}})}^{\frac{1}{b}})}^{b} = N^{b}

${({(\frac{1 + 2^{b}}{1 + 2^{- b}})}^{\frac{1}{b}})}^{b} = N^{b}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify

{({(\frac{1 + 2^{b}}{1 + 2^{- b}})}^{\frac{1}{b}})}^{b}

${({(\frac{1 + 2^{b}}{1 + 2^{- b}})}^{\frac{1}{b}})}^{b}$ .

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Step 3.2.1.1

Apply basic rules of exponents.

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Step 3.2.1.1.1

Apply the product rule to

\frac{1 + 2^{b}}{1 + 2^{- b}}

$\frac{1 + 2^{b}}{1 + 2^{- b}}$ .

{(\frac{{(1 + 2^{b})}^{\frac{1}{b}}}{{(1 + 2^{- b})}^{\frac{1}{b}}})}^{b} = N^{b}

${(\frac{{(1 + 2^{b})}^{\frac{1}{b}}}{{(1 + 2^{- b})}^{\frac{1}{b}}})}^{b} = N^{b}$

Step 3.2.1.1.2

Apply the product rule to

\frac{{(1 + 2^{b})}^{\frac{1}{b}}}{{(1 + 2^{- b})}^{\frac{1}{b}}}

$\frac{{(1 + 2^{b})}^{\frac{1}{b}}}{{(1 + 2^{- b})}^{\frac{1}{b}}}$ .

\frac{{({(1 + 2^{b})}^{\frac{1}{b}})}^{b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{{({(1 + 2^{b})}^{\frac{1}{b}})}^{b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

\frac{{({(1 + 2^{b})}^{\frac{1}{b}})}^{b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{{({(1 + 2^{b})}^{\frac{1}{b}})}^{b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

Step 3.2.1.2

Simplify the numerator.

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Step 3.2.1.2.1

Multiply the exponents in

{({(1 + 2^{b})}^{\frac{1}{b}})}^{b}

${({(1 + 2^{b})}^{\frac{1}{b}})}^{b}$ .

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Step 3.2.1.2.1.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

\frac{{(1 + 2^{b})}^{\frac{1}{b} b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{{(1 + 2^{b})}^{\frac{1}{b} b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

Step 3.2.1.2.1.2

Cancel the common factor of

b

$b$ .

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Step 3.2.1.2.1.2.1

Cancel the common factor.

\frac{{(1 + 2^{b})}^{\frac{1}{b} b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{{(1 + 2^{b})}^{\frac{1}{b} b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

Step 3.2.1.2.1.2.2

Rewrite the expression.

\frac{{(1 + 2^{b})}^{1}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{{(1 + 2^{b})}^{1}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

\frac{{(1 + 2^{b})}^{1}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{{(1 + 2^{b})}^{1}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

\frac{{(1 + 2^{b})}^{1}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{{(1 + 2^{b})}^{1}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

Step 3.2.1.2.2

Simplify.

\frac{1 + 2^{b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{1 + 2^{b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

\frac{1 + 2^{b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}

$\frac{1 + 2^{b}}{{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}} = N^{b}$

Step 3.2.1.3

Simplify the denominator.

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Step 3.2.1.3.1

Multiply the exponents in

{({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}

${({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}$ .

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Step 3.2.1.3.1.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{\frac{1}{b} b}} = N^{b}

$\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{\frac{1}{b} b}} = N^{b}$

Step 3.2.1.3.1.2

Cancel the common factor of

b

$b$ .

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Step 3.2.1.3.1.2.1

Cancel the common factor.

\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{\frac{1}{b} b}} = N^{b}

$\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{\frac{1}{b} b}} = N^{b}$

Step 3.2.1.3.1.2.2

Rewrite the expression.

\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{1}} = N^{b}

$\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{1}} = N^{b}$

\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{1}} = N^{b}

$\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{1}} = N^{b}$

\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{1}} = N^{b}

$\frac{1 + 2^{b}}{{(1 + 2^{- b})}^{1}} = N^{b}$

Step 3.2.1.3.2

Simplify.

\frac{1 + 2^{b}}{1 + 2^{- b}} = N^{b}