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  1. MHB Master
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    #1
    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.?


    Is everything correct so far?

    Could you give me a hint for the remaining $3$ properties? Do we use the previous properties?
    Last edited by mathmari; January 12th, 2020 at 07:47.

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  3. MHB Master
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    #2
    Assuming you are taking $ \displaystyle e^x$ as defined and then defining $ \displaystyle ln(x)$ as it's inverse function then, yes, those are correct.

  4. MHB Master
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    #3 Thread Author
    Ok!


    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?

  5. MHB Seeker
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    #4
    Quote Originally Posted by mathmari View Post
    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.

    Quote Originally Posted by mathmari View Post
    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.

    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:
    $$\ln(a^x\cdot a^y)=\ln(a^x)+\ln(a^y)=x\ln a+y\ln a=(x+y)\ln a=\ln(a^{x+y}) \implies a^x\cdot a^y=a^{x+y}$$

  6. MHB Master
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    #5
    I first learned exponentials an logarithms, like most people, learning about $ \displaystyle e^x$ first then having $ \displaystyle ln(x)$ defined as the inverse function to $ \displaystyle e^x$. But one can, as Klaas Van Aarsen suggests, define $ \displaystyle ln(x)$ as $ \displaystyle \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) $ \displaystyle 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, $ \displaystyle \lim_{x\to\infty} ln(x)= \infty$ and $ \displaystyle \lim_{x\to -\infty} ln(x)= -\infty$.

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

    6) $ \displaystyle ln(xy)= \int_1^{xy}\frac{1}{t}dt$. Let $ \displaystyle z= t/y$ so that $ \displaystyle t= yz$, $ \displaystyle dz=y dt$. When t= 1, z= 1/y and when t= xy, z= x. The integral becomes $ \displaystyle \int_{1/y}^x \frac{1}{yz}ydz= \int_{1/y}^x \frac{1}{z}dz$. We can write that as $ \displaystyle \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) $ \displaystyle ln(x^y)= \int_1^{x^y}\frac{1}{t}dt$. If y is not 0, let $ \displaystyle z= t^{1/y}$ so that $ \displaystyle t= z^y$ and [tex]dt= yz^{y-1}dz[/tez]. When t= 1, $ \displaystyle z= 1^{1/y}= 1$ and when $ \displaystyle t= x^y$, [tex]z= (x^y)^{1/y}= x[tex]. The integral becomes $ \displaystyle \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, [tex]x^0= 1[tex] and we already know that ln(1)= 0= 0(ln(x)). So $ \displaystyle ln(x^y)= y ln(x)$.

    We know that ln(x) is an increasing (so one-to-one) function from "positive real numbers" to "all real numbers" so it has an inverse function from "all real numbers" to "positive real numbers". Let "E(x)" 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:

    If E(x)= y then x= ln(y). If x is not 0 we can write that as $ \displaystyle 1= (1/x)ln(y)= ln(y^{1/x})$. Reversing, $ \displaystyle Exp(1)= y^{1/x}$ so that $ \displaystyle 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 "e" to be Exp(1) (so that ln(e)= 1) then the inverse function to f(x)= ln(x) is $ \displaystyle 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.

    Finally, if $ \displaystyle y= e^x$ then $ \displaystyle x= ln(y)$ so $ \displaystyle \frac{dx}{dy}= \frac{1}{y}$. Then $ \displaystyle \frac{dy}{dx}= \frac{1}{\frac{1}{y}}= y$. That is $ \displaystyle \frac{de^x}{dx}= e^x$.
    Last edited by Klaas van Aarsen; January 16th, 2020 at 08:29. Reason: Correct LaTeX Code

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