What is (4^{n+1})/((2^{n-1))^{n+1}} ?

The solution to (4^{n+1})/((2^{n-1))^{n+1}} is 2^{-n^2+2n+3}

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simplify (4^{n+1})/((2^{n-1))^{n+1}}

simplify

\frac{4 ^{n + 1}}{( 2 ^{n - 1} ) ^{n + 1}}

2^{- n^{2} + 2 n + 3}

Solution steps

\frac{4 ^{n + 1}}{( 2 ^{n - 1} ) ^{n + 1}}

Apply exponent rule:

(a^{b})^{c} = a^{b c}, a \geq 0

(2^{n - 1})^{n + 1} = 2^{(n - 1) (n + 1)}

= \frac{4 ^{n + 1}}{2 ^{(n - 1) (n + 1)}}

Factor

4^{n + 1} : 2^{2 (n + 1)}

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

Simplify

\frac{2 ^{2 (n + 1)}}{2 ^{(n - 1) (n + 1)}} : 2^{- n^{2} + 2 n + 3}

= 2^{- n^{2} + 2 n + 3}

\frac{x + 4}{2 x - 1} \leq 3

\frac{x + 2}{x - 4} < 1

- x^{2} + 4 x + 3 = ∣ x ∣ - 2

x^{2} + 5 x + \frac{25}{4}

2 lo g_{b} (q) + 8 lo g_{b} (t)

What is (4^{n+1})/((2^{n-1))^{n+1}} ?
The solution to (4^{n+1})/((2^{n-1))^{n+1}} is 2^{-n^2+2n+3}