# A Question about Defining Logarithms as integrals

1. Sep 12, 2011

### mahmoud2011

When Logarithm are Defined as integral and the Exponetial functions are defined to be its inverses , then What can prove ? or why ?

$a^n = \underbrace{a.a.a...a}_{n-times} : n\in N$

also Why we define rational exponents as roots?

I am sorry If my Question is silly.

IS my Question Clear ??

Last edited by a moderator: Sep 12, 2011
2. Sep 12, 2011

### mathman

It is completely unclear.

Last edited by a moderator: Sep 12, 2011
3. Sep 12, 2011

### mahmoud2011

I sorry , I mean When defining Exponential functions as inverses of Logarithms at which the natural logarithm is defined as integral , now Why the Exponential Functions result from this definition are same as Exponential functions we now in Elementary algebra , where after for example defining the base 2 Exponential function to be ,
$2^{x} = e\ ^{x.ln(2)}$

what make us sure that this Exponential function is the same as that we defined in Elementary algebra where for example , why when defining in this way we have , 2² = 4 = 2×2 ?

And my second question concerns in the rational non integer exponent is defined to be roots.more precisely,

$\sqrt[n]{a} = a^{\frac{1}{n}}$ where n is natural number , $n \geq 1$

Thanks .

Last edited by a moderator: Sep 12, 2011
4. Sep 12, 2011

### Staff: Mentor

Using the formula above to evaluate 23, we get
23 = e3 * ln(2) = eln(2)3 = 23 = 8, which is what you would expect from evaluating 23 directly.

The main reason is that $e\ ^{x.ln(2)} = e^{ln(2^x)} = 2^x$. The idea here is that e raised to the power ln(whatever) is whatever, as long as whatever > 0.
We know that $\sqrt{x}\sqrt{x} = x, \text{ for } x \geq 0$, so it's natural to associate the square root of a number with an exponent of 1/2.

x1/2*x1/2 = x1/2 + 1/2 = x1 = x, again supposing that x >= 0.

You can apply the same sort of reasoning for cube roots, fourth roots, and so on.

5. Sep 12, 2011

### mahmoud2011

Here is becomes my problem in defining the natural base exronential function e, we used rational exponents.

6. Sep 12, 2011

### AlephZero

When you define the "log function" as an integral, you need to prove that the function you defined actually behaves like a logarithm.

For example if you define a function L (note, I'm deliberately NOT calling it "log" or "ln", so I won't accidentally assume it has the same properties as logarithms)

$$L(x) = \int_1^x dt/t$$

Then you can prove basic things like $L(xy) = L(x) + L(y)$, straight from the definition:

$$L(xy) = \int_1^{xy} dt/t = \int_1^x dt/t + \int_x^{xy} dt/t$$

And by substituting u = xt, you can show that

$$\int_x^{xy} dt/t = \int_1^y dt/t = L(y)$$

So $L(xy) = L(x) + L(y)$.

I'm not sure exactly what you are asking about "rational exponents", but it follows directly from the definiton that $L(x^n) = nL(x)$, etc.

7. Sep 12, 2011

### mahmoud2011

Thanks,

When we define The natural logarithm we must keep in mind that we don't know any thing about the properties of exponents because we are going to proof them.now we Know that the inverse of ln(x) is exp(x) where

y=exp(x) if and only if ln(y)=x .............................. (1)

and we know

exp(ln (x) ) = x for all x>0
ln(exp(x)) = x for all x in R

in particular from previous
ln(0)=1
exp(1) = e

if r is rational number then ln(e^r) = rln(x) =r which implies exp(r)=e^r from (1)
and then we can define e^x even if x is irrational , the problem is here What makes e^r when r is rational (i.e How can we define the rational exponent before we go in this result )

8. Sep 12, 2011

### Staff: Mentor

What do you mean by this...
If you understand what ex means for an irrational number x, er for a rational number r is much simpler. You can always approximate an irrational number by using rational numbers that get closer and closer.

9. Sep 12, 2011

### mahmoud2011

I mean that when we put e to a rational exponent we must now what rational exponents first so how to define them how to know they exist ??

10. Sep 13, 2011

### HallsofIvy

We define rational exponents, r, of a number, a, by $a^r= e^{r ln(a)}$ of course. And we know such a number exists because we can show that the ln(x) function maps the set of all positive real numbers to the set of all real numbers: if y is any real number, there exist x such that $ln(x)= y$.

To see that, note that ln(x), defined as an integral, is continuous and differentiable on any interval of positive real numbers and, in particular, we can apply the "mean value theorem" to the interval [1, 2]: There exist c in that interval such that
$$\frac{ln(2)- ln(1)}{2- 1}= ln(2)= \frac{1}{c}$$.

But if $c\le 2$, then $ln(2)= \frac{1}{c}\ge 1/2$. That is important because we can now say that for any positive M, $ln(2^{2M})= 2Mln(2)\ge M$.
That is, ln(x) does NOT have an upper bound. And since its derivative is postive it is an increasing function: $\lim_{x\to\infty} ln(x)= \infty$. Further, since ln(1/x)= -ln(x), $\lim_{x\to 0} ln(x)= -\infty$. ln(x) maps the set of positive real number, one to one (because its derivative is always positive) onto the set of all real numbers and so has an inverse, exp(x), from the set of all real numbers to the set of all positive real numbers.

Now, here is what I think you are really asking about- How do we know that the inverse function, exp(x), to ln(x) really is "some number to the x power".

If y= exp(x), then x= ln(y). For x non-zero, we can divide by it to get 1= (1/x)ln(y)= ln(y^{1/x}).
Going back to the "exp" form, that says $y^{1/x}= ln(1)$ which is the same as $y= (ln(1))^x$. If x= 0, y must be 1 so $1= (ln(1))^0$ is still true.

That is, this "exp" function, defined as the inverse of the ln function really is some number to the x power. If we then define e to be ln(1), $exp(x)= e^x$.

Last edited by a moderator: Sep 13, 2011
11. Sep 13, 2011

### lurflurf

We can define exp and log any number of ways. If we fix one arbitrary constant and require the functions to be slightly well behaved there can be only one of each. Thus any exponential (logarithmic) function is the exponential (logarithmic) function.
If we define exp(r) for rational r we can define exp(x) for real x as
exp(x)=lim exp(r)
where r->x

12. Sep 13, 2011

### mahmoud2011

Do you mean exp(1) = e , ok

Thanks,I began to understand what I wanted , but This a part from I wanted to Know exactly .

When you write 1=ln(y^{1/x}) , here how to define y^1/x if 1/x is rational , and you defined first that a^r = e^{rln(a)} , that is when you define the rational expnonent , defined it depending that we know what is e^{rln(a)} which you defined using 1=ln(y^{1/x}) in the arguments. that was I am not understanding.

13. Sep 13, 2011

### lurflurf

^
often one defines
x^y:=exp(y log(x))
for this to work we must define exp and log
one common way is
log'(x)=1/x
log(exp(x))=x

but this is not the only way to do it.
There are many other ways.

14. Sep 13, 2011

### mahmoud2011

Please if you can write the other ways.Because in this way I can't see why for example a^2 = a.a

15. Sep 13, 2011

### lurflurf

we could define exp(x) as the only function such that
1)
exp(x)exp(y)=exp(x+y) for all real x and y
and
exp'(0)=1
2)
exp'(x)-exp(x)=0
and
exp(1)=e
3)
exp(x)=1+x+x^2/2+...+x^n/n!+...
(an infinite series)
4)
exp(x)=lim h->0 (1+x h)^(1/h)
5) define e then
exp(x)=e^x
5a)
if e^x is not preveously defined let
e^x=lim r->x e^r
where r is rational
5b)
define exp(x)=e^x for all x rational
require exp(x) be continuous

we could define log(x) as the only function such that
1)
log(xy)=log(x)+log(y) for all positive real x and y
and
log'(0)=1
1a) say you wish there were a function that made multiplication "like" addition to help with astronomy 400 years ago. Talk about Briggs and Napier.
1b)say homomorphism a lot during 1
2)
log'(x)=1/x
and
log(e)=1
2a)
make a big fuss over the distinction in 2) between antiderivatives and definite integrals
2b)Lie that people wished there were a name for the function that gives the area under a hyperbola, call it log.
3)
log(1+x)=x-x^2/2+...-(-1)^n x^n/n+...
(an infinite series)
4)
log(x)=lim h->0 (-1+x^h)/h

if one or the other is defined, define the other by
exp(log(x)=x

and on and on...

16. Sep 13, 2011

### mahmoud2011

Here is my problem that is how to define e^r when r is rational as I said .

I am sorry if I waste your time with me , and Thank you very much.

17. Sep 13, 2011

### lurflurf

to define e^r r rational
suppose we have first defined e
perhaps e=lim (1+1/n)^n
suppose r=a/b a an integer b a positive integer
we want
e^(a/b)=(e^a)^(1/b)=(e^(1/b))^a
first define e^a inductively by
e^(a+1)=e e^a
ie
e^0=1
e^1=1*e=e
e^2=e*e
e^3=e*e^2=e*e*e
and so on
supose a a positive integer
e^0=e^(a-a)=e^a*e^-a
so e^-a=1/e^a
supose a a positive integer
1=e^(a/a)=[e^(1/a)]^a
thus e^(1/a) must be the ath root of e
combining the above we can work out e^r for any rational r in terms of roots and integer powers of e

18. Sep 13, 2011

### mahmoud2011

But used one of the laws of exponents which are proved by defining logarithms as integrals first.

19. Sep 14, 2011

### lurflurf

Defining logarithms as integrals is only one possible way.
We can for example take the laws of exponents for rational numbers.
e^r*e^s=e^(r+s) r and s rational
Now if we want to define e^x for x real we require the functions to agree for rational values.
rationalexp(x)=realexp(x) when x is rational
what to do when x is not rational?
We would like exp well behaved, perhaps continuous.
Only one choice is possible.
If x is real and r is rational and r is close to x we must have e^r close to e^x.
There are an infinite number of rationals. So for each real x there is only one value for e^x that allows exp to be continuous.

20. Sep 14, 2011

### mahmoud2011

My Question is not concerned in defining irrational exponents from rational ones , My Question is how we can define the rational exponent as a root before Defining Logarithms as Integrals,Because when we defined e^x we used a previous knowledge of rational exponents so how to define them not using the laws.