O lvl EM: Indices

Question from http://www.sgforums.com/forums/2297/topics/313896

Use indices to solve the following:
3^y = 6^z show that z = (xy)/(x+y)

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Answer:

2^x = 6^z => 2^(1/z) = 6^(1/x) ------ (1)
3^y = 6^z => 3^(1/z) = 6^(1/y) ------ (2)

(1) * (2):
2^(1/z) * 3^(1/z) = 6^(1/x) * 6^(1/y)
6^(1/z) = 6^{(1/x + 1/y)}

thus

1/z = 1/x + 1/y

rearranging

z = (xy)/(x+y)

O lvl EM: Indices

Question from http://www.sgforums.com/forums/2297/topics/313742

(900 -p)^½ + 4(81-p)^½ = 50

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Answer:

let y = 81 - p

thus,

(819 + y)^½ + 4y^½ = 50
(819 + y)^½ = 50 - 4y^½

square both sides

819 + y = 2500 - 400 y^½ + 16y
15y - 400y^½ +1681 = 0

Using formula,

{ -b±√(b²- 4ac) } / 2a

y^½ = 5.23 or 21.44 (reject because y >0, (81-p)^½ = y^½ => y has to be positive)

Thus,

p = 81 - y = 81 - 5.23² = 53.65