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Question : 32 of 86

Marks: +1, -0

If

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

n

is equal to;

6
8
5
7 .

Solution:

Concept:

Use laws of indices to factor and solve for the exponent

n

.

Explanation:

Given

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

.

Rewrite

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

.

Then

2^{n-1} + 4 \cdot 2^{n-1} = 5 \cdot 2^{n-1} = 320

.

Divide both sides by 5:

2^{n-1} = 64

.

Since

64 = 2^6

, we have

2^{n-1} = 2^6

, so

n-1 = 6

.

Thus

n = 7

.

Answer:

n = 7

, which corresponds to option D (the fourth option).

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