# Proof

Let $$x,y$$ be real numbers and let $$a,b$$ be positive real numbers.

#### $$(i)$$ $$a^{x+y}=a^x\cdot a^y$$

It follows from the definition of the general power function and the functional equation of exponential function of general base.

#### $$(ii)$$ $$(a^x)^y=a^{x\cdot y}$$

The rule can be concluded as follows from the definition of exponential function of general base, and the definition of natural logarithm: Because $a^x=\exp_a(x)=\exp(x\ln(a))\Longleftrightarrow \ln(a^x)=x\ln a,$ we get $a^{xy}=\exp_a(xy)=\exp(yx\ln a)=\exp(y\ln(a^x))=(a^x)^y.$

#### $$(iii)$$ $$a^x\cdot b^x=(ab)^x$$

$\begin{array}{rcll} a^x\cdot b^x&=&\exp_a(x)\cdot\exp_b(x)&\text{definition of general power function}\\ &=&\exp(x\cdot \ln a)\cdot\exp(x\cdot \ln b)&\text{definition of exponential function of general base}\\ &=&\exp(x\cdot \ln a+ x\cdot \ln b)&\text{functional equation of exponential function}\\ &=&\exp(x\cdot (\ln a+ \ln b))&\text{distributivity law for real numbers}\\ &=&\exp(x\cdot (\ln (a\cdot b))&\text{functional equation of natural logarithm}\\ &=&\exp_{ab}(x)&\text{definition of exponential function of general base}\\ &=&(ab)^x&\text{definition of general power function}\\ \end{array}$

#### $$(iv)$$ $$a^{-x}=\frac 1{a^x}$$

This calculation rule follows from the definition of the general power function and the reciprocity of exponential function of general base.

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983