What is the answer to 2^{n-1}+2^{n+1}=320 ?

The answer to 2^{n-1}+2^{n+1}=320 is n=7

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2^{n-1}+2^{n+1}=320

2^{n - 1} + 2^{n + 1} = 320

n = 7

Solution steps

2^{n - 1} + 2^{n + 1} = 320

Factor

2^{n - 1} + 2^{n + 1} : 5 \cdot 2^{- 1 + n}

5 \cdot 2^{- 1 + n} = 320

Divide both sides by

5

2^{- 1 + n} = 64

Apply exponent rules

- 1 + n = 6

Solve

- 1 + n = 6 : n = 7

n = 7

3^{3 x - 1} = 2^{x + 3}

1 0^{x - 1} = 1000

3^{x + 1} = 4^{x - 1}

2^{x} = \frac{1}{32}

2^{x} + 4^{x} = 3

What is the answer to 2^{n-1}+2^{n+1}=320 ?
The answer to 2^{n-1}+2^{n+1}=320 is n=7
Solve for n: 2^{n-1}+2^{n+1}=320
The solution is n=7