- #1

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Is the integral of 1/x

ln x og log x?

ln x og log x?

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- Thread starter kasse
- Start date

- #1

- 366

- 0

Is the integral of 1/x

ln x og log x?

ln x og log x?

- #2

- 1,037

- 3

The integral is lnx.

- #3

- 529

- 1

Proof:

y=ln(x)

d/dx ln(x)= dy/dx

x=e^y

dx/dy=x

so dy/dx=1/x

y=ln(x)

d/dx ln(x)= dy/dx

x=e^y

dx/dy=x

so dy/dx=1/x

- #4

HallsofIvy

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Possible cause of the confusion: some texts, especially advanced texts, use "log" to mean the natural logarithm (ln) rather than the common logarithm (base 10). Common logarithms are seldom used now.

In general, since log_{a}(x)= ln(x)/ln(a),

[tex]\frac{d log_a(x)}{dx}= \frac{1}{x ln(a)}[/tex]

Of course, ln(e)= 1 so that reduces to the 1/x for the derivative of ln(x).

In general, since log

[tex]\frac{d log_a(x)}{dx}= \frac{1}{x ln(a)}[/tex]

Of course, ln(e)= 1 so that reduces to the 1/x for the derivative of ln(x).

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- #5

- 1,425

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Proof:

y=ln(x)

d/dx ln(x)= dy/dx

x=e^y

dx/dy=x

so dy/dx=1/x

That is not a proof at all. The natural number e is found with the definition of ln, the integral of 1/x. There is no "proof", it's a definition to start with. The only thing that needs to be proven is that ln behaves as a logarithmic function.

- #6

- 529

- 1

That is not a proof at all. The natural number e is found with the definition of ln, the integral of 1/x. There is no "proof", it's a definition to start with. The only thing that needs to be proven is that ln behaves as a logarithmic function.

Whatever dude.

It's a proof given the defn. d/dx e^x=e^x and defining ln to be the inverse operation to e^.

I agree that you could define ln(x) by its derivative. A check on Wikipedia shows both definitions.

If you define ln(x) from its derivative- it remains to be shown that e^ln(x)=x. The proof is essentially the same as I gave- working the other way.

- #7

Gib Z

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[tex]\int_1^x \frac{1}{t} dt = \ln x[/tex]. From this definition we can define its base, and we Euler decided to call it e. So e can be defined to be the number which fulfills the condition that

[tex]\int_1^e \frac{1}{x} dx = 1[/tex]

- #8

Gib Z

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Whatever dude.

It's a proof given the defn. d/dx e^x=e^x and defining ln to be the inverse operation to e^x.

That is eqivalent to the definition that e is the unique number that fulfills [tex]\lim_{h\to 0} \frac{e^h -1}{h} = 1 [/tex].

Definitons of e and Ln are not hard to make. The real challenge is to prove alternate definitons are in fact eqivalent. Wergs why makes things much easier.

- #9

- 529

- 1

[tex]\int_1^x \frac{1}{t} dt = \ln x[/tex]. From this definition we can define its base, and we Euler decided to call it e. So e can be defined to be the number which fulfills the condition that

[tex]\int_1^e \frac{1}{x} dx = 1[/tex]

That's fine. You have to define ln(x) somehow. I prefer to define it as the inverse operation to e^x, and then derive the integral of 1/x from there.

So, in fact, my full defn. would start from showing that e^x satisfies certain properties and working upwards- whereas you're starting from the other end and working backwords.

- #10

HallsofIvy

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The reason mathwonk says that defining ln x in terms of the integral is easier, and I agree, is that it avoids having to prove that

[tex]\lim_{x\rightarrow 0}\frac{a^x- a}{x}[/tex]

exists.

- #11

Gib Z

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...it avoids having to prove that

[tex]\lim_{x\rightarrow 0}\frac{a^x- a}{x}[/tex]

exists.

[tex]\lim_{h\to 0} \frac{e^{x+h} - e^x}{h}

= e^x \lim_{h\to 0} \frac{e^h -1}{h}[/tex]

Did you mean 1 or instead of a, or am i missing something?

- #12

HallsofIvy

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Yes, I meant

[tex]\lim_{x\rightarrow 0}\frac{a^x-1}{x}[/tex]

Thanks

[tex]\lim_{x\rightarrow 0}\frac{a^x-1}{x}[/tex]

Thanks

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