Given that 10^0.48 = x, 10^0.70 = y and x^z = y² then the value of z is close to = ?

$$\eqalign{ & {x^z}{\text{ = }}{y^2}{\text{ }} \cr & \Leftrightarrow {\left( {{{10}^{0.48}}} \right)^z} = {\left( {{{10}^{0.70}}} \right)^2} \cr & \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{\left( {2 \times 0.70} \right)}} = {10^{1.40}} \cr & \Leftrightarrow 0.48z = 1.40 \cr & \Leftrightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} \cr & \Leftrightarrow z = 2.9\left( {{\text{approx}}} \right) \cr} $$