MHB Properties of exponential/logarithm

[mathmari]

Hey!

I want to prove the following properties:

1. $\left (e^x\right )^y=e^{xy}$
2. $\ln (1)=0$
3. $\ln \left (x^y\right )=y\ln (x)$
4. $a^x\cdot a^y=a^{x+y}$ and $\frac{a^x}{a^y}=a^{x-y}$
5. $a^x\cdot b^x=\left (ab\right )^x$ and $\frac{a^x}{b^x}=\left (\frac{a}{b}\right )^x$
6. $\left (a^x\right )^y=a^{xy}$

I have done the following:

1. 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 $\displaystyle{\log \left (x^a\right )=a\cdot \log x}$.
   
2. We have that $e^0=1$. We apply the logarithm and we get $\displaystyle{\ln \left (e^0\right )=\ln (1)\Rightarrow 0\cdot \ln (e)=\ln (1)\Rightarrow 0=\ln (1)}$.
   
3. 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.? (Wondering)
   

Is everything correct so far? (Wondering)

Could you give me a hint for the remaining $3$ properties? Do we use the previous properties? (Wondering)

 

[HallsofIvy]

Assuming you are taking e^x as defined and then defining ln(x) as it's inverse function then, yes, those are correct.

 

[mathmari]

Ok! (Yes) For the remaining three properties I have done now the following:

4. We have that $a^x=e^{\ln \left (a^x\right )}$ and $a^y=e^{\ln \left (a^y\right )}$.

We multily the two terms and we get \begin{equation*}a^x\cdot a^y=e^{\ln \left (a^x\right )}\cdot e^{\ln \left (a^y\right )}=e^{\ln \left (a^x\right )+\ln \left (a^y\right )}=e^{x\cdot \ln \left (a\right )+y\cdot \ln \left (a\right )}=e^{(x+y)\cdot \ln \left (a\right )}=e^{ \ln \left (a^{x+y}\right )}=a^{x+y}\end{equation*}

For the second property we divide the terms above and we get \begin{equation*}\frac{a^x}{ a^y}=\frac{e^{\ln \left (a^x\right )}}{ e^{\ln \left (a^y\right )}}=e^{\ln \left (a^x\right )-\ln \left (a^y\right )}=e^{x\cdot \ln \left (a\right )-y\cdot \ln \left (a\right )}=e^{(x-y)\cdot \ln \left (a\right )}=e^{ \ln \left (a^{x-y}\right )}=a^{x-y}\end{equation*}
5. We multily the two terms and we get \begin{equation*}a^x\cdot b^x=e^{\ln \left (a^x\right )}\cdot e^{\ln \left (b^x\right )}=e^{\ln \left (a^x\right )+\ln \left (b^x\right )}=e^{x\cdot \ln \left (a\right )+x\cdot \ln \left (b\right )}=e^{x\cdot \left (\ln \left (a\right )+\ln \left (b\right )\right )}=e^{x\cdot \ln \left (ab\right )}=e^{ \ln \left (ab\right )^x}=\left (ab\right )^x\end{equation*}

For the second property we divide the terms above and we get \begin{equation*}\frac{a^x}{ b^x}=\frac{e^{\ln \left (a^x\right )}}{ e^{\ln \left (b^x\right )}}=e^{\ln \left (a^x\right )-\ln \left (b^x\right )}=e^{x\cdot \ln \left (a\right )-x\cdot \ln \left (b\right )}=e^{x\cdot \left (\ln \left (a\right )-\ln \left (b\right )\right )}=e^{x\cdot \ln \left (\frac{a}{b}\right )}=e^{ \ln \left (\frac{a}{b}\right )^x}=\left (\frac{a}{b}\right )^x\end{equation*}
6. We have that $a^x=e^{\ln \left (a^x\right )}$.

We get\begin{equation*}\left (a^x\right )^y=\left (e^{\ln \left (a^x\right )}\right )^y\overset{(1)}{=}e^{y\ln \left (a^x\right )}=e^{xy\ln \left (a\right )}=e^{\ln \left (a^{xy}\right )}=a^{xy}\end{equation*}
Are these proofs also correct? (Wondering)

 

[I like Serena]

3. $\ln \left (x^y\right )=y\ln (x)$

At 1. I used this rule, although I proved that here. Would there be also an other way to prove the property 1.?

We can do this one differently with the equivalent definition $\ln x \overset{\text{def}}{=} \int_1^x \frac{du}{u}$ and the substitution rule. (Thinking)

For the remaining three properties I have done now the following:

4. We have that $a^x=e^{\ln \left (a^x\right )}$ and $a^y=e^{\ln \left (a^y\right )}$.

We multily the two terms and we get \begin{equation*}a^x\cdot a^y=e^{\ln \left (a^x\right )}\cdot e^{\ln \left (a^y\right )}=e^{\ln \left (a^x\right )+\ln \left (a^y\right )}=e^{x\cdot \ln \left (a\right )+y\cdot \ln \left (a\right )}=e^{(x+y)\cdot \ln \left (a\right )}=e^{ \ln \left (a^{x+y}\right )}=a^{x+y}\end{equation*}

For the second property we divide the terms above and we get \begin{equation*}\frac{a^x}{ a^y}=\frac{e^{\ln \left (a^x\right )}}{ e^{\ln \left (a^y\right )}}=e^{\ln \left (a^x\right )-\ln \left (a^y\right )}=e^{x\cdot \ln \left (a\right )-y\cdot \ln \left (a\right )}=e^{(x-y)\cdot \ln \left (a\right )}=e^{ \ln \left (a^{x-y}\right )}=a^{x-y}\end{equation*}
5. We multily the two terms and we get \begin{equation*}a^x\cdot b^x=e^{\ln \left (a^x\right )}\cdot e^{\ln \left (b^x\right )}=e^{\ln \left (a^x\right )+\ln \left (b^x\right )}=e^{x\cdot \ln \left (a\right )+x\cdot \ln \left (b\right )}=e^{x\cdot \left (\ln \left (a\right )+\ln \left (b\right )\right )}=e^{x\cdot \ln \left (ab\right )}=e^{ \ln \left (ab\right )^x}=\left (ab\right )^x\end{equation*}

For the second property we divide the terms above and we get \begin{equation*}\frac{a^x}{ b^x}=\frac{e^{\ln \left (a^x\right )}}{ e^{\ln \left (b^x\right )}}=e^{\ln \left (a^x\right )-\ln \left (b^x\right )}=e^{x\cdot \ln \left (a\right )-x\cdot \ln \left (b\right )}=e^{x\cdot \left (\ln \left (a\right )-\ln \left (b\right )\right )}=e^{x\cdot \ln \left (\frac{a}{b}\right )}=e^{ \ln \left (\frac{a}{b}\right )^x}=\left (\frac{a}{b}\right )^x\end{equation*}
6. We have that $a^x=e^{\ln \left (a^x\right )}$.

We get\begin{equation*}\left (a^x\right )^y=\left (e^{\ln \left (a^x\right )}\right )^y\overset{(1)}{=}e^{y\ln \left (a^x\right )}=e^{xy\ln \left (a\right )}=e^{\ln \left (a^{xy}\right )}=a^{xy}\end{equation*}
Are these proofs also correct?

Yep. Correct. (Nod)

Btw, we might also use the properties $\ln(x\cdot y)=\ln x+\ln y$ and $\ln(x^y)=y\ln x$ and the fact that $\ln$ is bijective to find:
$$\ln(a^x\cdot a^y)=\ln(a^x)+\ln(a^y)=x\ln a+y\ln a=(x+y)\ln a=\ln(a^{x+y}) \implies a^x\cdot a^y=a^{x+y}$$
(Thinking)

 

[HallsofIvy]

I first learned exponentials an logarithms, like most people, learning about e^x first then having ln(x) defined as the inverse function to e^x. But one can, as Klaas Van Aarsen suggests, define ln(x) as \int_1^\infty \frac{1}{t}dt. From that we can immediately (well, the first time I worked on this it took a little longer!)

show​

that:

1) Since 1/x is continuous for all non-zero x but is not defined for x= 0, and the integral starts at t= 1> 0, ln(x) is defined for all positive x but is not defined for x 0 or negative.

2) Since ln(x) is defined as an integral, ln(x) is continuous and differentiable for all positive x and the derivative is 1/x.

3) ln(1)= \int_1^1 \frac{1}{t}dt= 0..

4) Since the derivative is positive, ln(x) is an increasing function. ln(x) is negative for x< 1 and positive for x>1 and, in fact, \lim_{x\to\infty} ln(x)= \infty and \lim_{x\to -\infty} ln(x)= -\infty.

5) ln(1/x)= \int_1^{1/x} \frac{1}{t}dt. Let y= 1/t so that t= 1/y and dt= -1/y^2 dy. When t= 1 y= 1/1= 1 and when t= 1/x y= x. So the integral becomes \int_1^x y\frac{-1}{y^2}dy= -\int_1^x \frac{1}{y}dy= -ln(x). So ln(1/x)= -ln(x).

6) ln(xy)= \int_1^{xy}\frac{1}{t}dt. Let z= t/y so that t= yz, dz=y dt. When t= 1, z= 1/y and when t= xy, z= x. The integral becomes \int_{1/y}^x \frac{1}{yz}ydz= \int_{1/y}^x \frac{1}{z}dz. We can write that as \int_{1/y}^1 \frac{1}{z}dz+ \int_{1}^x\frac{1}{z}dz= -\int_{1}^{1/y}\frac{1}{z}dz+ \int_1^x \frac{1}{z}dz= -(-ln(y))+ ln(x)= ln(x)+ ln(y). So ln(xy)= ln(x)+ ln(y).

7) ln(x^y)= \int_1^{x^y}\frac{1}{t}dt. If y is not 0, let z= t^{1/y} so that t= z^y and dt= yz^{y-1}dz[/tez]. When t= 1, z= 1^{1/y}= 1 and when t= x^y, z= (x^y)^{1/y}= x. The integral becomes \int_1^x\frac{1}{z^y}(yz^{y-1}dz= \int_1^x\frac{y}{z}dz= y\int_1^x\frac{1}{z}dz= yln(x). If y= 0, x^0= 1and we already know that ln(1)= 0= 0(ln(x)). So ln(x^y)= y ln(x).&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; We know that ln(x) is an increasing (so one-to-one) function from &amp;amp;amp;amp;quot;positive real numbers&amp;amp;amp;amp;quot; to &amp;amp;amp;amp;quot;all real numbers&amp;amp;amp;amp;quot; so it has an inverse function from &amp;amp;amp;amp;quot;all real numbers&amp;amp;amp;amp;quot; to &amp;amp;amp;amp;quot;positive real numbers&amp;amp;amp;amp;quot;. Let &amp;amp;amp;amp;quot;E(x)&amp;amp;amp;amp;quot; be that inverse function. We can prove a number of properties, such as E(x+y)= E(x)E(y) and E(x)^y= E(xy) but the most important are these:&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; If E(x)= y then x= ln(y). If x is not 0 we can write that as 1= (1/x)ln(y)= ln(y^{1/x}). Reversing, Exp(1)= y^{1/x} so that y= Exp(x)= (Exp(1))^x. That is, the inverse function to ln(x) really is just some number, Exp(1), to the x power. If we define &amp;amp;amp;amp;quot;e&amp;amp;amp;amp;quot; to be Exp(1) (so that ln(e)= 1) then the inverse function to f(x)= ln(x) is f^{-1}(x)= e^x. If x= 0 then, since ln(1)= 0, Exp(0)= 1 and any non-zero number, in particular e, to the 0 power is 1.&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Finally, if y= e^x then x= ln(y) so \frac{dx}{dy}= \frac{1}{y}. Then \frac{dy}{dx}= \frac{1}{\frac{1}{y}}= y. That is \frac{de^x}{dx}= e^x.