${\displaystyle {}{\begin{aligned}b^{x+x'}&=\exp \left((x+x')\ln b\right)\\&=\exp(x\ln b+x'\ln b)\\&=\exp \left(x\ln b\right)\cdot \exp \left(x'\ln b\right)\\&=b^{x}\cdot b^{x'}.\end{aligned}}}$

${\displaystyle {}b^{x}\cdot b^{-x}=b^{x-x}=b^{0}=\exp \left(0\right)=1\,,}$

deshalb sind ${\displaystyle {}b^{x}}$ und ${\displaystyle {}b^{-x}}$ invers zueinander.

${\displaystyle {}b>1}$

${\displaystyle {}x>0}$

${\displaystyle {}x\ln b>0\,}$

und damit ist nach Fakt  (5) ${\displaystyle {}b^{x}=\exp x\ln b>1}$.

${\displaystyle {}b>1}$

${\displaystyle {}\ln b>0}$

${\displaystyle {}x'>x}$

${\displaystyle {}x'\ln b>x\ln b}$

${\displaystyle {}b^{x'}=\exp \left(x'\ln b\right)>\exp \left(x\ln b\right)=b^{x}\,}$

aus Fakt  (6).

${\displaystyle {}b^{x}=\exp \left(x\ln b\right)\,}$

ist ${\displaystyle {}\ln b^{x}=x\ln b}$ und daher ist

${\displaystyle {}{\begin{aligned}(b^{x})^{x'}&=\exp \left(x'\cdot \ln \left(b^{x}\right)\right)\\&=\exp \left(x'\cdot x\cdot \ln \left(b\right)\right)\\&=b^{x\cdot x'}.\end{aligned}}}$

${\displaystyle {}{\begin{aligned}(ab)^{x}&=\exp \left(x\ln \left(ab\right)\right)\\&=\exp \left(x(\ln a+\ln b)\right)\\&=\exp \left(x\ln a+x\ln b\right)\\&=\exp \left(x\ln a\right)\cdot \exp \left(x\ln b\right)\\&=a^{x}\cdot b^{x}.\end{aligned}}}$