Solve for b N=b root of (1+2^b)/(1+2^(-b))

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

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

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

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

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[b]{\frac{1 + 2^{b}}{1 + 2^{- b}}}$ as ${(\frac{1 + 2^{b}}{1 + 2^{- b}})}^{\frac{1}{b}}$ .

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

Simplify the left side.

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

Apply basic rules of exponents.

Apply the product rule to $\frac{1 + 2^{b}}{1 + 2^{- b}}$ .

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

Apply the product rule to $\frac{{(1 + 2^{b})}^{\frac{1}{b}}}{{(1 + 2^{- b})}^{\frac{1}{b}}}$ .

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

Simplify the numerator.

Multiply the exponents in ${({(1 + 2^{b})}^{\frac{1}{b}})}^{b}$ .

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

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

Cancel the common factor of $b$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the denominator.

Multiply the exponents in ${({(1 + 2^{- b})}^{\frac{1}{b}})}^{b}$ .

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

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

Cancel the common factor of $b$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Solve for $b$ .

Take the log of both sides of the equation.

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

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

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

Expand $\ln (N^{b})$ by moving $b$ outside the logarithm.

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

Solve the equation for $b$ .

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Subtract $b \ln (N)$ from both sides of the equation.

$\ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) - b \ln (N) = 0$

Reorder $\ln (\frac{1 + 2^{b}}{1 + 2^{- b}})$ and $- b \ln (N)$ .

$- b \ln (N) + \ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) = 0$

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

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

The expanded equation is $\ln (1 + 2^{b}) - \ln (1 + 2^{- b}) = b \ln (N)$ .

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

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Subtract $b \ln (N)$ from both sides of the equation.

$\ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) - b \ln (N) = 0$

Reorder $\ln (\frac{1 + 2^{b}}{1 + 2^{- b}})$ and $- b \ln (N)$ .

$- b \ln (N) + \ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) = 0$

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

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

The expanded equation is $\ln (1 + 2^{b}) - \ln (1 + 2^{- b}) = b \ln (N)$ .

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

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Subtract $b \ln (N)$ from both sides of the equation.

$\ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) - b \ln (N) = 0$

Reorder $\ln (\frac{1 + 2^{b}}{1 + 2^{- b}})$ and $- b \ln (N)$ .

$- b \ln (N) + \ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) = 0$

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

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

The expanded equation is $\ln (1 + 2^{b}) - \ln (1 + 2^{- b}) = b \ln (N)$ .

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

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Subtract $b \ln (N)$ from both sides of the equation.

$\ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) - b \ln (N) = 0$

Reorder $\ln (\frac{1 + 2^{b}}{1 + 2^{- b}})$ and $- b \ln (N)$ .

$- b \ln (N) + \ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) = 0$

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

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

The expanded equation is $\ln (1 + 2^{b}) - \ln (1 + 2^{- b}) = b \ln (N)$ .

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

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Subtract $b \ln (N)$ from both sides of the equation.

$\ln (\frac{1 + 2^{b}}{1 + 2^{- b}}) - b \ln (N) = 0$

Solve for b N=b root of (1+2^b)/(1+2^(-b))

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