Question

Solve for x: 2^(x+1) + 2^(x-1) = 320

Option 1: x = 5
Option 2: x = 6
Option 3: x = 7
Option 4: x = 8

Answer 1

x = 7

Explanation:

consider the rules of exponents

•
`a^(m)` ×
`a^(n)` ⇔
`a^((m+n))`

•
`a^(-m)` ⇔
`(1)/(a^(m) )`

given

`2^((x+1))` +
`2^((x-1))` = 320 , then

`2^(x)` ×
`2^(1)` +
`2^(x)` ×
`2^(-1)` = 320 ← factor out
`2^(x)` from each term on the left side

`2^(x)` (
`2^(1)` +
`2^(-1)` ) = 320

`2^(x)` ( 2 +
`(1)/(2)` ) = 320

`2^(x)` ×
`(5)/(2)` = 320

multiply both sides by
`(2)/(5)`

`2^(x)` ×
`(5)/(2)` ×
`(2)/(5)` =
`(2)/(5)` × 320 , simplify both sides

`2^(x)` × 1 = 128 , that is

`2^(x)` = 128 [ note that 128 =
`2^(7)` ], so

`2^(x)` =
`2^(7)`

since bases on both sides are the same , both 2, then equate exponents

that is x = 7