Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to?

Answer: Option C

$$\begin{array}{rl}& 10^{0.48}=x\\ & 10^{0.70}=y\\ & \text{and }{x}^z=y^2\\ & \therefore {\left({10}^{0.48}\right)}^z={\left({10}^{0.70}\right)}^2\\ & \Rightarrow 10^{0.48z}=10^{1.40}\end{array}$$
If a^x = a^y, if base equal power are equal : (x = y)
$$\begin{array}{rl}& \therefore 0.48z=1.40\\ & \Rightarrow z=\frac{140}{48}\\ & \Rightarrow z=\frac{35}{12}\\ & \Rightarrow z=2.9\end{array}$$