Can natural logs help solve equations involving e?

[Andrew123]

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

 

[Santa1]

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?

 

[Integral]

A revealing relationship is this:

$$\int _1 ^e \frac {dx} x = 1$$

 

[n_bourbaki]

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).

 

[HallsofIvy]

In general, the function log_a(x) is defined as the inverse function to a^x 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 $log_a(a^x)= x$ and $a^{log_a(x)}$.

 

[maze]

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.

 

[gmax137]

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).

 

[Werg22]

As Integral pointed out, the relationship

$$\int _1 ^e \frac {dx} x = 1$$

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

$$\int _1 ^x \frac {dt} t$$

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.

 

[HallsofIvy]

As Integral pointed out, the relationship

$$\int _1 ^e \frac {dx} x = 1$$

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

$$\int _1 ^x \frac {dt} t$$

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
$$ln(x)= \int_1^x \frac{dt}{t}$$
Then, for y any real number
$$ln(x^y)= \int_1^{x^y}\frac{dt}{t}$$
If y is not 0, define $u= t^{1/y}$ so that $t= u^y$ and $dt= y u^{y-1} du$
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]<br /> <br /> Of course, if y=0, then x^y= 1 and ln(1)= 0. So even for y= 0, ln(x^y)= ln(1)= 0= y ln(x).<br /> <br /> Thus for all <b>real</b> x, we have ln(x^y)= y ln(x)[/tex]

 

[HallsofIvy]

By the way, after defining
[tex]ln(x)= \int_1^x \frac{dt}{t}[/itex] <br /> 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.<br /> <br /> That means that we can apply the mean value theorem on the interval [1, 2]:<br /> $$\frac{ln(2)- ln(1)}{2- 1}= \frac{1}{t}$$<br /> for some t between 0 and 1.<br /> Of course, the left hand side is just ln(2) so we have <br /> $$ln(2)= \frac{1}{t}$$<br /> for $1\le t\le 2$. But that means $1/2\le 1/t\le 1$ or <br /> [tex]ln(2)\ge 1/2[/itex]<br /> <br /> Now, given any positive number X, we have <br /> $$ln(2^{2X})= 2Xln(2)\ge (2X)(1/2)= X$$<br /> That is, that ln(x) is <b>unbounded</b> and, since it is increasing,<br /> [tex]\lim_{x\rightarrow \infty} ln(x)= \infty[/itex]<br /> Further, since ln(x^-1)= -ln(x), <br /> [tex]\lim_{x\rightarrow 0} ln(x)= -\infty[/itex]<br /> <br /> That tells us that ln(x) is a <b>one-to-one</b> function from R⁺ to R and so has an inverse function, exp(x), from R to R⁺.<br /> <br /> 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(y^1/x). Going back to exp form, <br /> y^1/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.<br /> <br /> That shows that exp(x), defined in this rather convoluted manner, is, in fact, <b>some number to the x power</b>. If we now define e to be exp(1) (that is, the number such that ln(e)= 1) then we have that exp(x)= e^x.[/tex][/tex][/tex][/tex]