MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

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}.$1 Igor's answer is probably better but I'll observe that we want to interpolate between$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 that would work at both ends.