Question

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

Answer 1

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)