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fitzpratik
Given that \(10^\frac{48}{100}\) = x, \(10^\frac{7}{10}\) = y and\(x^z\) = \(y^2\), then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36


\(y = 10^{\frac{7}{10}}\)

\(y^2 = 10^{\frac{7}{10} * 2} = 10^{\frac{7}{5}}\)

\(x = 10^{\frac{48}{100}} = 10^{\frac{12}{25}}\)

\(x^z = 10^{\frac{12}{25}}^z = 10^{\frac{7}{5}} = y^2\)

So \((\frac{12}{25})z = \frac{7}{5}\)

\(z = \frac{35}{12} = 2.9\)

Answer (C)
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We get

10^(48z/100) =10^14/10

Equating exponents (since we are in base 10) gives 48z/100 =14/10
then 480z=1400
and z=1400/480. 480*3=1440, therefore the answer is a little less than 3. C
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