Solve for n (b^5)/(b^n)=1/b

$\frac{b^{5}}{b^{n}} = \frac{1}{b}$

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

$\ln (\frac{b^{5}}{b^{n}}) = \ln (\frac{1}{b})$

Expand the left side.

Rewrite $\ln (\frac{b^{5}}{b^{n}})$ as $\ln (b^{5}) - \ln (b^{n})$ .

$\ln (b^{5}) - \ln (b^{n}) = \ln (\frac{1}{b})$

Expand $\ln (b^{5})$ by moving $5$ outside the logarithm.

$5 \ln (b) - \ln (b^{n}) = \ln (\frac{1}{b})$

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

$5 \ln (b) - (n \ln (b)) = \ln (\frac{1}{b})$

Remove parentheses.

$5 \ln (b) - n \ln (b) = \ln (\frac{1}{b})$

Reorder $5 \ln (b)$ and $- n \ln (b)$ .

$- n \ln (b) + 5 \ln (b) = \ln (\frac{1}{b})$

Move all the terms containing a logarithm to the left side of the equation.

$- n \ln (b) + 5 \ln (b) - \ln (\frac{1}{b}) = 0$

Move all terms not containing $n$ to the right side of the equation.

Subtract $5 \ln (b)$ from both sides of the equation.

$- n \ln (b) - \ln (\frac{1}{b}) = - 5 \ln (b)$

Add $\ln (\frac{1}{b})$ to both sides of the equation.

$- n \ln (b) = - 5 \ln (b) + \ln (\frac{1}{b})$

Divide each term in $- n \ln (b) = - 5 \ln (b) + \ln (\frac{1}{b})$ by $- 1 \ln (b)$ and simplify.

Divide each term in $- n \ln (b) = - 5 \ln (b) + \ln (\frac{1}{b})$ by $- 1 \ln (b)$ .

$\frac{- n \ln (b)}{- 1 \ln (b)} = \frac{- 5 \ln (b)}{- 1 \ln (b)} + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{n \ln (b)}{\ln (b)} = \frac{- 5 \ln (b)}{- 1 \ln (b)} + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

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

Cancel the common factor.

$\frac{n \ln (b)}{\ln (b)} = \frac{- 5 \ln (b)}{- 1 \ln (b)} + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

Divide $n$ by $1$ .

$n = \frac{- 5 \ln (b)}{- 1 \ln (b)} + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

Simplify the right side.

Simplify each term.

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

Cancel the common factor.

$n = \frac{- 5 \ln (b)}{- 1 \ln (b)} + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

Rewrite the expression.

$n = \frac{- 5}{- 1} + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

Move the negative one from the denominator of $\frac{- 5}{- 1}$ .

$n = - 1 \cdot - 5 + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

Rewrite $- 1 \cdot - 5$ as $- - 5$ .

$n = - - 5 + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

Multiply $- 1$ by $- 5$ .

$n = 5 + \frac{\ln (\frac{1}{b})}{- 1 \ln (b)}$

Move the negative in front of the fraction.

$n = 5 - \frac{\ln (\frac{1}{b})}{\ln (b)}$

Solve for n (b^5)/(b^n)=1/b

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