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

Answer: Option C

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