What is the answer to f(n)=(w^{n+1})/(w^{2n)} ?

The answer to f(n)=(w^{n+1})/(w^{2n)} is n=-(ln(f(x))-ln(w))/(ln(w))

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f(n)=(w^{n+1})/(w^{2n)}

f (n) = \frac{w ^{n + 1}}{w ^{2 n}}

n = - \frac{ln ( f ( x ) ) - ln ( w )}{ln ( w )}

Solution steps

f (n) = \frac{w ^{n + 1}}{w ^{2 n}}

Simplify

\frac{w ^{n + 1}}{w ^{2 n}} : w^{- n + 1}

f (n) = w^{- n + 1}

Solve

f (n) = w^{- n + 1} : n = - \frac{ln ( f ( x ) ) - ln ( w )}{ln ( w )}

n = - \frac{ln ( f ( x ) ) - ln ( w )}{ln ( w )}

\frac{x ^{2} - 1}{2 x - 4} = 0

\frac{( ( 2 n - 1 ) ! )}{( ( 2 n - 3 ) ! )} = 120

(\frac{x + 1}{x})^{2} + 5 (\frac{x + 1}{x}) + 4 = 0

\frac{- 5 + 3}{4 ∣ 1 - x ∣} = \frac{7}{3}

\frac{( x + 1 )}{x} + ∣ x + 1 ∣ = \frac{( ( x + 1 ) ^{2} )}{∣ x ∣}

What is the answer to f(n)=(w^{n+1})/(w^{2n)} ?
The answer to f(n)=(w^{n+1})/(w^{2n)} is n=-(ln(f(x))-ln(w))/(ln(w))