Understanding exponentiation and logarithms together

[mcastillo356]

TL;DR Summary

A quote from the maths textbook Calculus, by Robert A. Adams and Christopher Essex, and a graph I should conceptually connect, but do not.

Hi PF

The logarithm is the inverse function to exponentiation. But, focusing on exponentiation, here comes the graph:


And, next, the quote
An exponential function is a function of the form $f(x)=a^x$, where the base $a$ is a positive constant and the exponent $x$ is the variable.
Observe that $a^x>0$ for all $a>0$ and all real $x$ and that
If $a>1$, then $\lim_{x\rightarrow -\infty}{a^x=0}$ and $\lim_{x\rightarrow \infty}{a^x=\infty}$
If $0<a<1$, then $\lim_{x\rightarrow -\infty}{a^x=\infty}$ and $\lim_{x\rightarrow \infty}{a^x=0}$

The graph of $y=a^x$ has the x-axis as an horizontal asymptote if $a\neq 1$. It is asymptotic on the left (as $x\rightarrow{-\infty}$) if $a>1$ and on the right (as $x\rightarrow{\infty}$) if $0<a<1$

In my opinion, it is asymptotic on the left (as $x\rightarrow{-\infty}$) if $a>0$ and on the right (as $x\rightarrow{\infty}$) if $a>0$. This remark doesn't contradict the texbook, but I think it is rather acccurate.

Greetings!

 

[anuttarasammyak]

Two graphs of
$$y=a^x$$
$$y = \log_a x$$
are symmetric wrt line y=x. That would support your observation.

 

[pasmith]

In my opinion, it is asymptotic on the left (as $x\rightarrow{-\infty}$) if $a>0$ and on the right (as $x\rightarrow{\infty}$) if $a>0$. This remark doesn't contradict the texbook, but I think it is rather acccurate.

The case $a = 1$ is the constant function 1. This does not have the x-axis as a horizontal asymptote in either direction; it can only be said to have the line $y = 1$ as a horizontal asympytote.

Note that $a^x = (1/a)^{-x}$, so the graph of $a^x$ for $0 < a < 1$ is the reflection in the y-axis of the graph of $(1/a)^x$, where necessarily $(1/a) > 1$.

 

[mcastillo356]

Hi, @anuttarasammyak, there is no support in absolute to my observation. I've been wrong at all:

There is no logic in what I proposed to state. I'm writing maybe too fast, but, what is the sense in saying things like "when $a>0$, it is asymtotic on the left as $x\rightarrow{-\infty}$"?.

Still have to think how to explain the reason why my original post is not logical. I need some time.

@pasmith, I've also read #3 post rapidly, but now I must outline my mistake's cause.

Greetings!

 

[anuttarasammyak]

I am not sure if I get you. Another observation is
$$a^{-x}=\frac{1}{a^x}$$
As for $a^x$ and $a^{-x}$, when one limit is infnitely small another limit is infinitely large.

 

[mcastillo356]

Hi, PF

Two graphs of
$$y=a^x$$
$$y = \log_a x$$
are symmetric wrt line y=x. That would support your observation.

Hi, @anuttarasammyak, Geogebra has helped me to understand your quote. I think there is still $a^0=1$. Certainly, if $a\neq 1$, for all $a>0$, $y=\log_a x$ and $y=a^x$ are symmetric with respect to $y=x$. Recalling one of the properties of the inverse functions, $y=f^{-1} (x)\Leftrightarrow{x=f(y)}$, which in this case means that the graph of $y=\log_a x\Leftrightarrow{y=a^x}$. The fact quoted above is among the properties pointed out, and therefore, a support to my initial guess.

The case $a = 1$ is the constant function 1. This does not have the x-axis as a horizontal asymptote in either direction; it can only be said to have the line $y = 1$ as a horizontal asympytote.

I will quote Wikipedia: "For curves given by the graph of a function $y=f(x)$, horizontal asymptotes are horizontal lines that the graph of the function approaches as $x$ tends to $+\infty$ or $-\infty$." Therefore, $y=1$ is, in my opinion, part of $y=a^x$ behaviour, instead of an horizontal asymptote.

Note that $a^x = (1/a)^{-x}$, so the graph of $a^x$ for $0 < a < 1$ is the reflection in the y-axis of the graph of $(1/a)^x$, where necessarily $(1/a) > 1$.

Thanks a lot!

 

[Mark44]

Recalling one of the properties of the inverse functions, $y=f^{-1} (x)\Leftrightarrow{x=f(y)}$, which in this case means that the graph of $y=\log_a x\Leftrightarrow{y=a^x}$.

No, the latter equivalency is not true. The two equations in this equivalency produce two different graphs. These graphs are symmetric about the line y = x, but they aren't the same graph, so the equations are not equivalent. Thecorrected second equivalency should read $y=\log_a x \Leftrightarrow x = a^y$.

The $\Leftrightarrow$ symbol means that every pair of numbers x and y on the graph of the first equation is also on the graph of the second equation. That also means that the graphs of $y = \log_a(x)$ and $x = a^y$ are exactly the same.

The fact quoted above is among the properties pointed out, and therefore, a support to my initial guess.

I don't see how a false statement lends support to your initial guess.

 

[mcastillo356]

No, the latter equivalency is not true. The two equations in this equivalency produce two different graphs. These graphs are symmetric about the line y = x, but they aren't the same graph, so the equations are not equivalent. The corrected second equivalency should read $y=\log_a x \Leftrightarrow x = a^y$.

Yes

The $\Leftrightarrow$ symbol means that every pair of numbers x and y on the graph of the first equation is also on the graph of the second equation. That also means that the graphs of $y = \log_a(x)$ and $x = a^y$ are exactly the same.

Definite.

Let's go further in the texbook; right after entering the exponential function, introduces logarithms this way:
Logarithms
The function $f(x)=a^x$ is a one-to-one function provided that $a>0$ and $a\neq 1$. Therefore, $f$ has an inverse which we call logartithmic function.
DEFINITION 5
If $a>0$ and $a\neq 1$, the function $\log_a x$, called the logartithm of $x$ to the base $a$, is the inverse of the one-to-one function $a^x$:
$$y=\log_a x \quad \Leftrightarrow{\quad{x=a^y},\qquad{(a>0,\quad{a\neq 1})}}$$
Since $a^x$ has domain $(-\infty,\infty)$, $\log_a x$ has range $(-\infty,\infty)$. Since $a^x$ has range $(0,\infty)$, $\log_a x$ has domain $(0,\infty)$. Since $a^x$ and $\log_a x$ are inverse functions, the following cancellation identities hold:
$$\log_a (a^x)=x\qquad{\text{for all real}}\,x\qquad{\text{and}}\qquad{a^{\log_a x}=x}\quad{\text{for all}\quad{x>0}}$$

The graphs of some typical logarithmic functions are shown at the right of the Figure 3.8(b) (attached in the original post). They all pass through the point $(1,0)$. Each graph is the reflection in the line $y=x$ of the corresponding exponential graph in the original post's figure.

I don't see how a false statement lends support to your initial guess.

True quote. I've checked it with an example, using Geogebra


Now, let's revisit my initial guess. I was interested on examining the texbook and a commentary of my own.

TL;DR Summary: A quote from the maths textbook Calculus, by Robert A. Adams and Christopher Essex, and a graph I should conceptually connect, but do not.

The logarithm is the inverse function to exponentiation. But, focusing on exponentiation, here comes the graph:
View attachment 333768

In my opinion, it is asymptotic on the left (as $x\rightarrow{-\infty}$) if $a>0$ and on the right (as $x\rightarrow{\infty}$) if $a>0$. This remark doesn't contradict the texbook, but I think it is rather acccurate.

This quote is right. But this other

Hi, @anuttarasammyak, Geogebra has helped me to understand your quote. I think there is still $a^0=1$. Certainly, if $a\neq 1$, for all $a>0$, $y=\log_a x$ and $y=a^x$ are symmetric with respect to $y=x$. Recalling one of the properties of the inverse functions, $y=f^{-1} (x)\Leftrightarrow{x=f(y)}$, which in this case means that the graph of $y=\log_a x\Leftrightarrow{y=a^x}$. The fact quoted above is among the properties pointed out, and therefore, a support to my initial guess.

Belongs to my own vintage. Some weird rabbit I took out of my hat, based on the fact that the simmetry of the logarithmic function and the exponential function (this latter not like Wikipedia understands, being Euler's number the function's base, but any positive real number) with respect $y=x$, supports my opinion that it is possible to qualify, or comment on, the textbook, just the way I do it in this thread's original post.

Greetings!

 

[mcastillo356]

The case $a = 1$ is the constant function 1. This does not have the x-axis as a horizontal asymptote in either direction; it can only be said to have the line $y = 1$ as a horizontal asympytote.

Note that $a^x = (1/a)^{-x}$, so the graph of $a^x$ for $0 < a < 1$ is the reflection in the y-axis of the graph of $(1/a)^x$, where necessarily $(1/a) > 1$.

Hi, shouldn't it be "where neccesarily $(1/a)>0$". Did you mean $(1/a)^x>1$?
Greetings!

 

[pasmith]

Hi, shouldn't it be "where neccesarily $(1/a)>0$". Did you mean $(1/a)^x>1$?
Greetings!

I meant exactly what I said.

This is in the context of your assertion

In my opinion, it is asymptotic on the left (as $x\rightarrow{-\infty}$) if $a>0$ and on the right (as $x\rightarrow{\infty}$) if $a>0$. This remark doesn't contradict the texbook, but I think it is rather acccurate.

Firstly in both cases you require $a > 0$; I assume one of them is meant to be $a < 0$, but if $a < 0$ then $a^x$ is not necessarily real for non-integer $x$, as for example with $a = -1$ and $x = \frac12$.

My point, however, is that the behaviour depends on wether $a > 1 > 0$ or $0 < a < 1$, and we have the symmetry between the two cases which I noted in my post, with the diving case $a = 1$ being the horizontal line $y = 1$.

 

[mcastillo356]

Hi

Firstly in both cases you require $a > 0$;

Yes

I assume one of them is meant to be $a < 0$, but if $a < 0$ then $a^x$ is not necessarily real for non-integer $x$, as for example with $a = -1$ and $x = \frac12$.

I don't understand. I really don't know.

My point, however, is that the behaviour depends on wether $a > 1 > 0$ or $0 < a < 1$, and we have the symmetry between the two cases which I noted in my post, with the diving case $a = 1$ being the horizontal line $y = 1$.

I think so. When taking limits.
Greetings!

 

[Mark44]

Firstly in both cases you require $a > 0$; I assume one of them is meant to be $a < 0$, but if $a < 0$ then $a^x$ is not necessarily real for non-integer $x$, as for example with $a = -1$ and $x = \frac12$.

I don't understand. I really don't know.

@mcastillo356, you completely missed the point that @pasmith made; i.e., when a (the base) is negative, not when the exponent is negative. In the two graphs you included, a = 5 in both graphs. The graphs were of $y = 5^{-x}$ and $y = 5^x$.

@pasmith -- I find it to be less effort to use wrap LaTeX expressions with a pair of ## tags rather than typing out the itex tags.

 

[mcastillo356]

@mcastillo356, you completely missed the point that @pasmith made; i.e., when a (the base) is negative, not when the exponent is negative. In the two graphs you included, a = 5 in both graphs. The graphs were of $y = 5^{-x}$ and $y = 5^x$.

Thanks! My Geogebra file was a poor lonely example; some kind of decontextualized stuff.

The case $a = 1$ is the constant function 1. This does not have the x-axis as a horizontal asymptote in either direction; it can only be said to have the line $y = 1$ as a horizontal asympytote.

Note that $a^x = (1/a)^{-x}$, so the graph of $a^x$ for $0 < a < 1$ is the reflection in the y-axis of the graph of $(1/a)^x$, where necessarily $(1/a) > 1$.

This is the reality, not my narrow attempts to suit my vain intentions.

Thanks!