Solve for x given that {64}^(2x - 1) = {32}^(x + 1) ​

Question

Solve for x given that

`{64}^(2x - 1) = {32}^(x + 1)`
​

Answer 1

x =
`(11)/(7)`

Explanation:

note that 64 =
`2^(6)` and 32 =
`2^(5)` , thus

`(2^(6)) ^(2x-1)` =
`(2^(5)) ^(x+1)` , that is

`2^(6(2x-1))` =
`2^(5(x+1))`

`2^(12x-6)` =
`2^(5x+5)`

Since the bases on both sides are equal, then equate the exponents

12x - 6 = 5x + 5 ( subtract 5x from both sides )

7x - 6 = 5 ( add 6 to both sides )

7x = 11 ( divide both sides by 7 )

x =
`(11)/(7)`