A Question about Defining Logarithms as integrals

[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 ??

 

[mathman]

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 ??

It is completely unclear.

 

[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 .

 

[Mark44]

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 ,
$2^{x} = e\ ^{x.ln(2)}$

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

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 ?

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.

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 .

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.

x^1/2*x^1/2 = x^{1/2 + 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.

 

[mahmoud2011]

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.

x^1/2*x^1/2 = x^{1/2 + 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.

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

 

[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.

 

[mahmoud2011]

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 definition that $L(x^n) = nL(x)$, etc.

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 )

 

[Mark44]

What do you mean by this...

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 )

If you understand what e^x means for an irrational number x, e^r for a rational number r is much simpler. You can always approximate an irrational number by using rational numbers that get closer and closer.

 

[mahmoud2011]

What do you mean by this...


If you understand what e^x means for an irrational number x, e^r for a rational number r is much simpler. You can always approximate an irrational number by using rational numbers that get closer and closer.

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 ??

 

[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), $exp(x)= e^x$.

 

[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

 

[mahmoud2011]

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), $exp(x)= e^x$.

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.

 

[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.

 

[mahmoud2011]

^
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.

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

 

[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...

 

[mahmoud2011]

we could define exp(x) as the only function such that
5a)
if e^x is not preveously defined let
e^x=lim r->x e^r
where r is rational

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.

 

[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

 

[mahmoud2011]

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

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

 

[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.

 

[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.

 

[HallsofIvy]

As I said before, you don't have to define rational or irrational exponents if you define $e^x$ as the inverse function to the logarithm. Since ln(x) maps a the set of positive real numbers to the set of all real numbers so if r is any real number, integer, rational, or irrational, there exist x such that log(x)= r and so $e^x= r$. However, if you don't want to define the exponential function as the inverse to the natural logarithm, you can do the following:
1) Define $a^n$, for a any positive real number and n a positive integer, to be (a*a*a*...*a) n times. That is, a is multiplied n times.

From that, we can see that $(a^n)(a^m)= a^{m+n}$: in $a^ma^n$ we have n "copies" of a multiplied by m "copies" of a which is a total of m+n "copies" of a multiplied together.

Also, it is easy to see that $(a^n)^m= a^{nm}$: think of each $a^n$ as a row of n "copies" of a, with each of m "copies" of $a^n$ written one under the the other- that gives m rows of n "copies" of a for a total of mn "copies" of a multiplied together.

Now, those two, $a^na^m= a^{n+m}$ and $(a^n)^m= a^{mn}$ are very nice formulas so we define a^x for other kinds of numbers in a way that make those still true.

If n= 0 then $a^0a^m= a^{0+m}= a^m$. Since $a^m$, for m a positive integer is never 0, we can divide both sides by $a^m$ to get $a^0= 1$. That is, in order that "$a^ma^n= a^{m+n}$ still be true, we must define $a^0= 1$ for any positive real number, a.

Similarly, for n any positive integer, -n is negative and we want to have $a^na^{-n}= a^{n-n}= a^0= 1$. Again, $a^n$ is never 0 so we can divide both sides by $a^n$ to get $a^{-n}= 1/a^n$. That is, to have $a^ma^b= a^{m+n}$, we must define $a^{-n}= 1/a^n$.

For n any positive integer, 1/n exists and n(1/n)= 1. So if $(a^{1/n})^n= a^{n(1/n)}= a$ we must have $a^{1/n}$ an nth root of a. That is, to have $(a^m)^n= a^{mn}$ even for m or n of fraction of the form 1/k for some integer k, we must define $a^{1/n}= \sqrt[n]{a}$, the principle nth root of a.

Any rational number is of the form m/n for m an integer and n a positive integer. From $(a^m)^n= a^{mn}$, we have $a^{m/n}= (a^m)^{1/n}= \sqrt[n]{a^m}$.

That defines $a^r$ for r any rational number. You can then define $a^x$ as lurflurf says.

However, I want to reiterate that if you "define logartithms as integrals", you can then define $e^x$ for x any real number as the inverse function to ln(x). It is NOT necessary, as you seem to believe, to define rational exponents as roots.

$e^{1/2}$ is the value of x such that ln(x) =1/2- and we know that value exists. Nothing more is necessary. You can then, as I showed in my first response, show that exp(x) (defined as the inverse function to ln(x)) is the x power of some number.

 

[mahmoud2011]

As I said before, you don't have to define rational or irrational exponents if you define $e^x$ as the inverse function to the logarithm. Since ln(x) maps a the set of positive real numbers to the set of all real numbers so if r is any real number, integer, rational, or irrational, there exist x such that log(x)= r and so $e^x= r$. However, if you don't want to define the exponential function as the inverse to the natural logarithm, you can do the following:
1) Define $a^n$, for a any positive real number and n a positive integer, to be (a*a*a*...*a) n times. That is, a is multiplied n times.

From that, we can see that $(a^n)(a^m)= a^{m+n}$: in $a^ma^n$ we have n "copies" of a multiplied by m "copies" of a which is a total of m+n "copies" of a multiplied together.

Also, it is easy to see that $(a^n)^m= a^{nm}$: think of each $a^n$ as a row of n "copies" of a, with each of m "copies" of $a^n$ written one under the the other- that gives m rows of n "copies" of a for a total of mn "copies" of a multiplied together.

Now, those two, $a^na^m= a^{n+m}$ and $(a^n)^m= a^{mn}$ are very nice formulas so we define a^x for other kinds of numbers in a way that make those still true.

If n= 0 then $a^0a^m= a^{0+m}= a^m$. Since $a^m$, for m a positive integer is never 0, we can divide both sides by $a^m$ to get $a^0= 1$. That is, in order that "$a^ma^n= a^{m+n}$ still be true, we must define $a^0= 1$ for any positive real number, a.

Similarly, for n any positive integer, -n is negative and we want to have $a^na^{-n}= a^{n-n}= a^0= 1$. Again, $a^n$ is never 0 so we can divide both sides by $a^n$ to get $a^{-n}= 1/a^n$. That is, to have $a^ma^b= a^{m+n}$, we must define $a^{-n}= 1/a^n$.

For n any positive integer, 1/n exists and n(1/n)= 1. So if $(a^{1/n})^n= a^{n(1/n)}= a$ we must have $a^{1/n}$ an nth root of a. That is, to have $(a^m)^n= a^{mn}$ even for m or n of fraction of the form 1/k for some integer k, we must define $a^{1/n}= \sqrt[n]{a}$, the principle nth root of a.

Any rational number is of the form m/n for m an integer and n a positive integer. From $(a^m)^n= a^{mn}$, we have $a^{m/n}= (a^m)^{1/n}= \sqrt[n]{a^m}$.

That defines $a^r$ for r any rational number. You can then define $a^x$ as lurflurf says.

However, I want to reiterate that if you "define logartithms as integrals", you can then define $e^x$ for x any real number as the inverse function to ln(x). It is NOT necessary, as you seem to believe, to define rational exponents as roots.

$e^{1/2}$ is the value of x such that ln(x) =1/2- and we know that value exists. Nothing more is necessary. You can then, as I showed in my first response, show that exp(x) (defined as the inverse function to ln(x)) is the x power of some number.

Thank you very much I can express my thanks in words.
But please another final question is, when we solve for zeros of function and construct the sign diagram which is guaranteed by intermediate value so how we are sure that there are no zeros except for those expressed on the sign diagram . For example for exponential function. e^x - 2= 0 ,we know that e^x = 2 if and only if ln(e^x) =ln(2) because of one to one property
then x=ln(2)

and in other many and many other functions which we find there zeros by factorizing first and find zeros is the zero-factor theorem is the only thing which guarantee that we have exactly all zeros . My problem is that I want to know every proof in mathematics with the right concept.

 

[HallsofIvy]

In general, we can't. In the case of $e^x= 2$, we know there is only a single solution because we know that $e^x$ is a "one-to-one" function. And we know that because we know that the derivative of $e^x$ is $e^x$ which is positive for all x. Since the derivative is positive for all x, the function is strictly increasing and so we never have $e^{x_1}= e^{x_2}$ for $x_1\ne x_2$.

But, for general functions, even polynomials, there no "simple" way of determining how many zeroes there may be in a given function (except, as you say, if f(a)< 0 and f(b)>0 there must be at least one x such that f(x)= 0 between a and b).

 

[mahmoud2011]

so why when constructing the sign diagram we place the sign over the interval although we are not sure that It has no more zeros.In polynomials we can prove that the real zeros we obtain is the only real and the same with rational 's , radicals and with some exponential functions and of course trig. functions right?

 

[HallsofIvy]

What "sign diagram" are you talking about? The sign diagrams I know are either signs of the function when we know the zeroes of the function or signs of the the derivative when we know the zeros of the derivatives.

 

[mahmoud2011]

What "sign diagram" are you talking about? The sign diagrams I know are either signs of the function when we know the zeroes of the function or signs of the the derivative when we know the zeros of the derivatives.

Yes this is what I am talking about i place zeros and put (0) above them and use test values to detremine the sign of intervals also I place the excluded values for domain with "?" above them .

 

[mahmoud2011]

Yes this is what I am talking about i place zeros and put (0) above them and use test values to detremine the sign of intervals also I place the excluded values for domain with "?" above them .

Please help me at this point.for example when I have this simple equation x² - 2 = 0, we have the solution $\pm \sqrt{2}$, now is the factor theorem is what guarantees that these are the only solutions . now if I have the solutions of f(x) = 0 , must I prove that my solutions are the only real solutions in order to construct the sign diagram of the function or what ?

 

[HallsofIvy]

No! As you just said the fact that you can factor $x^2- 2= (x- \sqrt{2})(x+\sqrt{2})$ is enough to prove that only $x= \sqrt{2}$ and $x= -\sqrt{2}$ are roots.

 

[mahmoud2011]

^is that means that every time I construct the sign diagram , I must prove that my real solutions are exactly the only real solutions for the equation.?

 

[HallsofIvy]

Yes, sometimes you actually have to do the algebra!

 

[mahmoud2011]

Thanks