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

Answer: Option C

$$\begin{array}{rl}& x^z=y^2\iff 10^{\left(0.48z\right)}=10^{2\times 0.70}=10^{1.40}\\ & \Rightarrow 0.48z=1.40\\ & \Rightarrow z=\frac{140}{48}=\frac{35}{12}=2.9(\text{approx})\end{array}$$