Can natural logs help solve equations involving e?

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Of course, we could also have said that ln(e)= 2, or -7, or any other number. It is only the "ln(e)= 1" that matters.)In summary, e^(lnx) = x because that is the definition of ln x as the inverse of e^x. The integral of 1/x ln(x) is ln(x) + C because ln is the inverse of e and the integral of 1/x is ln(x). The relationship \int _1 ^e \frac {dx} x = 1 reveals the connection between e and ln. The book "e the story of a number" by Eli Maor explains the importance of e and logs. The function ln
  • #1
Andrew123
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Why does e ^ (lnx) = x? where ln is log base e. Why is the integral of 1/x ln(x) and not log(x) ie log base 10 x? Is there some good information somewhere explainined this please? Cheers
 
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  • #2
e^(lnx) = x because that's the way lnx is defined, the integral of 1/x is lnx + C because (lnx)' = 1/x, can you do (log_{10}(x))' and see what the big difference is?
 
  • #3
A revealing relationship is this:

[tex] \int _1 ^e \frac {dx} x = 1 [/tex]
 
  • #4
Andrew123 said:
Why does e ^ (lnx) = x?

Because that is its definition.

Why is the integral of 1/x ln(x)

Because ln is the inverse to e, and e^x is defined to be the unique (up to constants) function satisfying the differential equation f'(x)=f(x).
 
  • #5
In general, the function loga(x) is defined as the inverse function to ax and, of course, two functions, f and g, are "inverse" to one another if and only if f(g(x)= x and g(f(x))= x. Therefore, it is always true that [itex]log_a(a^x)= x[/itex] and [itex]a^{log_a(x)}[/itex].
 
  • #6
You can also show that d/dx ln(x) = 1/x from the limit definition of derivative. You have to bust out the epsilon-delta definition of limits though. The key points along the proof are realizing that 1+h/x = e^(h/x) + O( (h/x)^2 ), and using the fact that ln(x) <= x-1.
 
  • #7
If you are really interested, look for this book:
"e the story of a number" By Eli Maor. It explains a lot of the "e" stuff and logs and why we care about the natural logs (base e).
 
  • #8
As Integral pointed out, the relationship

[tex]
\int _1 ^e \frac {dx} x = 1
[/tex]

is an important one in elucidating this question. Usually, we define log x as

[tex]
\int _1 ^x \frac {dt} t
[/tex]

The constant e can be defined as the limit of (1 + 1/n)^n and it can be shown that log e = 1. That said, the equation log (a^b) = b * log a can be derived purely from the definition of log as an integral*. Then follows that log (e^x) = x*log e = x. This process shows that e^x is the inverse function of log x. And so e^log x = x *** Technically, I know only of such a development that assumes b is rational and expands to all real numbers later in the program, but assuming this is valid for all real b's still conveys the main idea.

** If f and g are inverses of one another, f(g(x)) = x.
 
  • #9
Werg22 said:
As Integral pointed out, the relationship

[tex]
\int _1 ^e \frac {dx} x = 1
[/tex]

is an important one in elucidating this question. Usually, we define log x as

[tex]
\int _1 ^x \frac {dt} t
[/tex]

The constant e can be defined as the limit of (1 + 1/n)^n and it can be shown that log e = 1. That said, the equation log (a^b) = b * log a can be derived purely from the definition of log as an integral*. Then follows that log (e^x) = x*log e = x. This process shows that e^x is the inverse function of log x. And so e^log x = x **


* Technically, I know only of such a development that assumes b is rational and expands to all real numbers later in the program, but assuming this is valid for all real b's still conveys the main idea.


Define
[tex]ln(x)= \int_1^x \frac{dt}{t}[/tex]
Then, for y any real number
[tex]ln(x^y)= \int_1^{x^y}\frac{dt}{t}[/tex]
If y is not 0, define [itex]u= t^{1/y}[/itex] so that [itex]t= u^y[/itex] and [itex]dt= y u^{y-1} du[/itex]
Then the integral becomes
[tex]\int_1^x \frac{y u^{y-1}du}{u^y}= y\int_1^x \frac{du}{u}= y ln(x)[/itex]

Of course, if y=0, then xy= 1 and ln(1)= 0. So even for y= 0, ln(x^y)= ln(1)= 0= y ln(x).

Thus for all real x, we have ln(xy)= y ln(x)
 
  • #10
By the way, after defining
[tex]ln(x)= \int_1^x \frac{dt}{t}[/itex]
It is easy to see that the derivative is 1/x which is positive for all positive x: that is, ln(x), defined for all positive x, is an increasing function. Since it is defined as an integral it is obviously continuous and, in fact, differentiable for all positive x.

That means that we can apply the mean value theorem on the interval [1, 2]:
[tex]\frac{ln(2)- ln(1)}{2- 1}= \frac{1}{t}[/tex]
for some t between 0 and 1.
Of course, the left hand side is just ln(2) so we have
[tex]ln(2)= \frac{1}{t}[/tex]
for [itex]1\le t\le 2[/itex]. But that means [itex]1/2\le 1/t\le 1[/itex] or
[tex]ln(2)\ge 1/2[/itex]

Now, given any positive number X, we have
[tex]ln(2^{2X})= 2Xln(2)\ge (2X)(1/2)= X[/tex]
That is, that ln(x) is unbounded and, since it is increasing,
[tex]\lim_{x\rightarrow \infty} ln(x)= \infty[/itex]
Further, since ln(x-1)= -ln(x),
[tex]\lim_{x\rightarrow 0} ln(x)= -\infty[/itex]

That tells us that ln(x) is a one-to-one function from R+ to R and so has an inverse function, exp(x), from R to R+.

But since they are inverse functions, y= exp(x) is the same as x= ln(y). Now, if x is not 0, we can write 1= (1/x)ln(y)= ln(y1/x). Going back to exp form,
y1/x= exp(1) so y= exp(1)x. Of course, if x= 0, y= exp(0)= 1 (because ln(1)= 0) and it is still true that 1= 1= exp(1)0 no matter what exp(1) is.

That shows that exp(x), defined in this rather convoluted manner, is, in fact, some number to the x power. If we now define e to be exp(1) (that is, the number such that ln(e)= 1) then we have that exp(x)= ex.
 

1. How do natural logs help solve equations involving e?

Natural logs, or ln, are the inverse of the exponential function e^x. This means that taking the natural log of both sides of an equation involving e allows us to isolate the variable and solve for it.

2. Can natural logs be used to solve any equation involving e?

Yes, natural logs can be used to solve any equation that involves the variable being raised to a power of e. However, they may not always be the most efficient method of solving the equation.

3. Are there any special rules or properties for using natural logs to solve equations involving e?

Yes, there are a few rules to keep in mind when using natural logs to solve equations involving e. These include the property ln(ab) = ln(a) + ln(b) and the rule for solving ln(e^x) = x.

4. Can natural logs be used to solve equations with other bases besides e?

Yes, natural logs can be used to solve equations with any base. However, the equation must be rewritten in terms of e first. For example, to solve an equation with base 10, we would use the rule ln(x) = ln(10^x).

5. Is it always necessary to use natural logs to solve equations involving e?

No, it is not always necessary to use natural logs to solve equations involving e. Other methods, such as substitution or algebraic manipulation, may be more efficient in certain cases.

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