If the eccentricity of the ellipse $\frac{{\mathrm{x}}^2}{{\left(\log \mathrm{a}\right)}^2}+\frac{{\mathrm{y}}^2}{{\left(\log \mathrm{b}\right)}^2}=1 \left(\mathrm{a}>\mathrm{b}>0, \mathrm{a}, \mathrm{b}\ne 1\right) \mathrm{is} \frac{1}{\sqrt{2}}$and 'e' be the eccentricity of the hyperbola $\frac{{\mathrm{x}}^2}{{\left({\log }_{\mathrm{b}}\mathrm{a}\right)}^2}-{\mathrm{y}}^2=1, \mathrm{then} {\mathrm{e}}^2$ is greater than (where $\log \mathrm{x}=\ln \mathrm{x})$