# Properties of exponential/logarithm

#### mathmari

##### Well-known member
MHB Site Helper
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:

#### HallsofIvy

##### Well-known member
MHB Math Helper
Assuming you are taking $$e^x$$ as defined and then defining $$ln(x)$$ as it's inverse function then, yes, those are correct.

#### mathmari

##### Well-known member
MHB Site Helper
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? #### Klaas van Aarsen

##### MHB Seeker
Staff member
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. 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:
$$\ln(a^x\cdot a^y)=\ln(a^x)+\ln(a^y)=x\ln a+y\ln a=(x+y)\ln a=\ln(a^{x+y}) \implies a^x\cdot a^y=a^{x+y}$$ #### HallsofIvy

##### Well-known member
MHB Math Helper
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 $$dt= yz^{y-1}dz[/tez]. When t= 1, [tex]z= 1^{1/y}= 1$$ and when $$t= x^y$$, $$z= (x^y)^{1/y}= x[tex]. The integral becomes [tex]\int_1^x\frac{1}{z^y}(yz^{y-1}dz= \int_1^x\frac{y}{z}dz= y\int_1^x\frac{1}{z}dz= yln(x)$$. If y= 0, $$x^0= 1[tex] and we already know that ln(1)= 0= 0(ln(x)). So [tex]ln(x^y)= y ln(x)$$.

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 $$1= (1/x)ln(y)= ln(y^{1/x})$$. 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 "e" 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.

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 $$\frac{de^x}{dx}= e^x$$.

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