Igor's answer is probably better but I'll observe that we want to interpolate between $e^{\ln{x}}$ at $t=0$ and and $e^x$ at $t=1.$ Furthermore $\int x^{-1}dx=\ln(x)+C$ and $\int x^0 dx=x+C.$
Rearranging a bit, since $\displaystyle \lim_{t \rightarrow 0^{+}}\frac{x^t-1}{t}=\ln{x},$ I am tempted by $$e^{\frac{x^t-1}{t}}=\sqrt[t]{e^{x^t-1}}.$$ That does not quite work at $t=1$ since it becomes $e^{x-1}.$ However it is also true that $\displaystyle \lim_{t \ rightarrow 0^{+}}\frac{x^t-1+t^2}{t}=\ln{x}\$ and $\ \displaystyle \lim_{t \rightarrow 0^{+}}\frac{x^t-1+t^{2-t}}{t}=\ln{x}$ so something like 0^{+}}\frac{(x+t)^t-1}{t}=\ln{x}$ and that would work works at both ends .as do $\frac{x^t-1+t^2}{t}\ $ and $\frac{x^t-1+t^{2-t}}{t}.$
