Question
Given that `10^0.48 = x, 10^0.70 = y and x^z = y^2 `, then the value of z is close to :
Answer: Option C
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`x^z = y^2 hArr (10^0.48)^z = (10^0.70)^2 hArr 10^(0.48z) = 10^(2 xx -.70) = 10^1.40`
`hArr` 0.48z = 1.40 ` hArr z = 140/48 = 35/12 ` = 2.9 (approx.)
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