How do you solve 1000 (1+x) ^12=2000

Solve for x over the real numbers:

1000 (x + 1) ^12 = 2000

Divide both sides by 1000:

(x + 1) ^12 = 2

Take the square root of both sides:

(x + 1) ^6 = sqrt (2) or (x + 1) ^6 = - sqrt (2)

Take the square root of both sides:

(x + 1) ^3 = 2^ (1/4) or (x + 1) ^3 = - 2^ (1/4) or (x + 1) ^6 = - sqrt (2)

Take cube roots of both sides:

x + 1 = 2^ (1/12) or (x + 1) ^3 = - 2^ (1/4) or (x + 1) ^6 = - sqrt (2)

Subtract 1 from both sides:

x = 2^ (1/12) - 1 or (x + 1) ^3 = - 2^ (1/4) or (x + 1) ^6 = - sqrt (2)

Take cube roots of both sides:

x = 2^ (1/12) - 1 or x + 1 = - 2^ (1/12) or (x + 1) ^6 = - sqrt (2)

Subtract 1 from both sides:

x = 2^ (1/12) - 1 or x = - 1 - 2^ (1/12) or (x + 1) ^6 = - sqrt (2)

(x + 1) ^6 = - sqrt (2) has no solution since for all x on the real line, (x + 1) ^6 = (x + 1) ^6 >=0 and - sqrt (2) <0:

Answer: x = 2^ (1/12) - 1 or x = - 1 - 2^ (1/12)