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Mathematics
Spencer Horn
12 June, 14:22
How do you solve 1000 (1+x) ^12=2000
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Spookey
12 June, 17:42
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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)
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