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Given that: 100.48 = x and 100.7 = y and xz = y2, find the value of z.

(a) \(2\frac{11}{12}\)

(b) \(\frac 49\)

(c) \(1\frac 1{48}\)

(d) \(\frac{48}{49}\)

2 Answers

+1 vote
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Best answer

Correct option is (a) \(2\frac{11}{12}\)

xz = y2 ⟺10(0.48z) = 10(2 × 0.70) = 101.40

⇒ 0.48z = 1.40

⇒ \(z = \frac{140}{48} = \frac{35}{12} = 2.9\) (approx)

+1 vote
by (14.8k points)

Another way to solve this problem is to use logarithms.

log(100.48) = log(x)     log(100.7) = log(y)   and  log(xz) = log(y2)

So: 0.48 = log(x)     0.7 = log(y)   and  z log(x) = 2 log(y)

Substituting: (0.48) z = (0.7)(2)   ==>   z = 35/12  

Correct answer is: (a) 2 11/12

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