If n is a positive integer such that 2^{n-2} + 2^{n-1} + 2^n + 2^{n+1} = 480, what is the … — GRE General Test

If $n$ is a positive integer such that $2^{n-2} + 2^{n-1} + 2^n + 2^{n+1} = 480$ , what is the value of $n$ ?

Answer: 7

The value of

n

is 7.

Factoring out

2^{n-2}

converts the sum into

2^{n-2}(1 + 2 + 4 + 8) = 15 \cdot 2^{n-2} = 480

. Dividing 480 by 15 gives

2^{n-2} = 32

. Since

32 = 2^5

, setting

n - 2 = 5

yields

n = 7

Factor out the lowest power of 2, which is

2^{n-2}

, from all terms on the left side of the equation.

2^{n-2}(1 + 2 + 4 + 8) = 480

Factoring out a common exponential term simplifies the addition of powers into a product of a single exponential term and a constant.

Evaluate the constant factor inside the parentheses and solve for the exponential expression

2^{n-2}

15 \cdot 2^{n-2} = 480 \implies 2^{n-2} = 32

Summing

1 + 2 + 4 + 8

yields 15. Dividing both sides by 15 isolates the base-2 term.

Express 32 as a power with base 2 and equate the exponents.

2^{n-2} = 2^5 \implies n - 2 = 5 \implies n = 7

Since

32 = 2^5

and the bases are identical, the exponents must be equal.

Factoring sum of exponential terms with common bases

Question