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

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