1

Natural logarithm is defined as:

$\ln(Y) = x$

Which can be also written as:

$e^x = y$

Now the problem is, to solve the above equation for x you would need to use logarithm, unless the base can be set to be the same. The solution is circular. In other words the logarithm solution to the exponential equation problem is not the final solution at all, as we just take the logs as given.

The question is, how do the calculators solve the equation for x and thus provide the log functions which can be used? Where does the final solution to the problem come from? Is it just trial and error or is there a formula which can be used behind it?

Dole
  • 2,653
  • They are inverses of each other, much like + and - or $\times$ and $\div$... – danimal Jun 11 '15 at 14:21
  • 2
    You shouldn't use lower-case $y$ and capital $Y$ interchangeably. It is not unusual that those would be used to represent two different things in the same problem. ${}\qquad{}$ – Michael Hardy Jun 11 '15 at 14:21
  • Well, you know that $e$ is a number, right? Around $2.718\dots$. So $e^x$ is just $2.718\dots$ to the $x$ power. As for how calculators do it… – Akiva Weinberger Jun 11 '15 at 14:22
  • …it can be proven that $e^x$ is equal to the infinite sum $1+\dfrac x{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dotsb$ for all $x$ (where $!$ means factorial).So, calculators can just compute that infinite sum to however many terms as they want (more terms means more accuracy). Alternatively, it's possible they have a table built-in where they can look up what $e^x$ is for at least some values of $x$. – Akiva Weinberger Jun 11 '15 at 14:25
  • It's an interesting exercise to prove that $e^xe^x=e^{2x}$ using that infinite series. (Because of the factorials, the series for $e^x$ is sometimes used in combinatorics.) – Akiva Weinberger Jun 11 '15 at 14:29
  • @columbus8myhw Thanks, this will do! It's the maclaurin series expansion for those interested. – Dole Jun 11 '15 at 14:40
  • @columbus8myhw : That series may converge fast if $-1<x<0$, but if I wanted to compute $e^5$ by hand, I might look for somewhat more nuanced methods. – Michael Hardy Jun 11 '15 at 15:04
  • @MichaelHardy Hm, you're right. Do you happen to know what these "nuanced methods" are? (Or do you have a link?) – Akiva Weinberger Jun 11 '15 at 15:17

3 Answers3

2

"Solve" is a word that gets used by non-mathematicians as a catch-all word -- something to use when they don't know what word to use. So does "equation".

One does not "solve" a function. One may solve an equation.

If you solve the equation $\ln y = x$ for $y$, you get $y=e^x$.

If you solve the equation $y=e^x$ for $x$, you get $x = \ln y$.

As far as numerical computations go, there are efficient algorithms for computing values of the exponential and logarithmic functions.

Michael Hardy
  • 1
  • So is there a solution to the function specified for x that does not involve the use of logs or trial and error? – Dole Jun 11 '15 at 14:30
  • 1
    There are algorithms for finding numerical values of the function that do not consist of taking logarithms or doing trial and error. But as I said, I wouldn't use the word "solution" in that way. – Michael Hardy Jun 11 '15 at 14:44
0

The definition of logarithm $\log_a(y)$ is indeed the solution to $a^x=y$.

However, the solution is not circular, since it clearly follows rules of arithmetic. One could say that it is "cheating" to "define" a solution rather than to really "arrive" into one. However, one can show with numerical methods that $\log_a(y)$ exists as a real and when applied, gives the correct result.

Addressing your question, calculators use CAS (Computer Algebra System) to evaluate the equation and arrive into the solution, much like pen and paper solution would. The algorithm goes trough the nature of the equation, handles the proper operations, and goes trough the equation again.

Michael Hardy
  • 1
Eemil Wallin
  • 693
  • I changed $log_a(y)$ to $\log_a(y)$, coded as \log_a(y). That not only prevents italicization but also provides proper spacing before and after $\log$ in expressions like $x\log_a y$. It is standard usage. ${}\qquad{}$ – Michael Hardy Jun 12 '15 at 15:50
  • Oh, I've just started learning the basics of LaTeX, thank you for the tip. – Eemil Wallin Jun 12 '15 at 16:04
0

Yes! The creation of the exponential and logarithmic functions are artificial in the sense that they are inverses of each other. So when Wikipedia defines the exponential function as the inverse of the natural log, you should feel a bit slighted.

$$(1) \quad e^x=\lim_{n \to \infty} \left(1+{1 \over n} \right)^{n \cdot x}$$

What does this say? Well, it says that if a number increases $(1+{1 \over n})$ percent per ${1 \over n}$ second as n approaches infinity, then growing by this amount for x seconds yields $e^x$. There are better sources available. Look here and at this.

For the $\ln(x)$ the most common definition is $$\ln(x)=\int_1^x {1 \over x} \ dx$$

This means the derivative of $\ln(x)$ is ${ 1 \over x}$. If we say the natural log has an inverse function $G(x)$ we arrive at. $$G^{'}(x)={1 \over {f^{'}(G(x))}}={1 \over {(G(x))^{-1}}}=G(x)$$

It's not difficult to show (1) has a derivative equal to itself and thus $e^x$ is indeed the inverse of $\ln(x)$.

How do you approximate these functions values? You can derive a series for $e^x$ fairly easily. If you say 1 equals $e^x$ you'll see that the derivative of 1 is 0, that's wrong, so we need to correct that bad approximation. We'll add $x$. Now we see the derivative is 1 however now we see that we have to add ${{x^2} \over 2}$ so the derivative is correct. We keep doing this and end up with.

$$e^x=1+x+{{x^2} \over {2!}}+{{x^3} \over {3!}}...$$

I won't prove it, but here's an expansion for $\ln(x)$

$$\ln(x)=x-{{x^2} \over 2}+{{x^3} \over 3}-{{x^4} \over 4}...$$

Note the similarities between the two series. Hope that helps!

Zach466920
  • 8,341