I want to prove the following properties:
- $\left (e^x\right )^y=e^{xy}$
- $\ln (1)=0$
- $\ln \left (x^y\right )=y\ln (x)$
- $a^x\cdot a^y=a^{x+y}$ and $\frac{a^x}{a^y}=a^{x-y}$
- $a^x\cdot b^x=\left (ab\right )^x$ and $\frac{a^x}{b^x}=\left (\frac{a}{b}\right )^x$
- $\left (a^x\right )^y=a^{xy}$
I have done the following:
We have that $\displaystyle{\left (e^x\right )^y=e^{\ln \left (e^x\right )^y}=e^{y\cdot \ln e^x}=e^{y\cdot x}=e^{xy}}$
We used the rule $\displaystyle{\log \left (x^a\right )=a\cdot \log x}$.
We have that $e^0=1$. We apply the logarithm and we get $\displaystyle{\ln \left (e^0\right )=\ln (1)\Rightarrow 0\cdot \ln (e)=\ln (1)\Rightarrow 0=\ln (1)}$.
We have that $e^{\ln x}=x$. We raise the equation to $n$ and we get $\left (e^{\ln x}\right )^y=x^y \Rightarrow e^{y\cdot \ln x}=x^y$. We take the logarithm of the equation and we get $\ln \left (e^{y\cdot \ln x}\right )=\ln \left (x^y\right ) \Rightarrow y\cdot \ln x=\ln \left (x^y\right )$.
At 1. I used this rule, although I proved that here. Would there be also an other way to prove the property 1.?
Is everything correct so far?
Could you give me a hint for the remaining 3 properties? Do we use the previous properties?
For 4:
Let's say $a^x\cdot a^y = b$ for a number b. Taking the logarithms of both sides give : $\ln(a^x\cdot a^y) = \ln(b).$ Using the product rule for logarithms we get : $\ln(a^x) + \ln(a^y) = \ln(b)$ , and from the power rule we get : $(x+y)\cdot \ln(a) = \ln(b) $. Raising both sides to the e-th power give : $b = e^{(x+y)\cdot \ln(a)} = a^{x+y}$.