What is ((3^{n+1})/(2^n)+(2^{n+1})/(3^n)-5)^{-1} ?

The solution to ((3^{n+1})/(2^n)+(2^{n+1})/(3^n)-5)^{-1} is 1/(((3^{n+1))/(2^n)+(2^{n+1})/(3^n)-5)}

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simplify ((3^{n+1})/(2^n)+(2^{n+1})/(3^n)-5)^{-1}

simplify

(\frac{3 ^{n + 1}}{2 ^{n}} + \frac{2 ^{n + 1}}{3 ^{n}} - 5)^{- 1}

\frac{1}{( \frac{3 ^{n + 1}}{2 ^{n}} + \frac{2 ^{n + 1}}{3 ^{n}} - 5 )}

Solution steps

(\frac{3 ^{n + 1}}{2 ^{n}} + \frac{2 ^{n + 1}}{3 ^{n}} - 5)^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{( \frac{3 ^{n + 1}}{2 ^{n}} + \frac{2 ^{n + 1}}{3 ^{n}} - 5 )}

- lo g_{10} (1.8) 1 0^{- 4}

\frac{2 ^{3}}{3 !} \cdot e^{- 2}

3 x (3 x)

What is ((3^{n+1})/(2^n)+(2^{n+1})/(3^n)-5)^{-1} ?
The solution to ((3^{n+1})/(2^n)+(2^{n+1})/(3^n)-5)^{-1} is 1/(((3^{n+1))/(2^n)+(2^{n+1})/(3^n)-5)}