Discussion
Given that \(10^{0.48}=x\), \(10^{0.70}=y\) and \(x^{z}=y^{2}\), then the value of z is close to:
Answer: C
\(x^{z}=y^{2}\) \(\Leftrightarrow 10^{(0.48z)}=10^{2\times 0.70}=10^{1.40}\)
\(\Rightarrow 0.48z=1.40\)
\(\Rightarrow z=\frac{140}{48}=\frac{35}{12}=2.9(approx.)\)
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