Proof: "lnx ≠ x" (please check!)

[Andrew Roberts]

Hi,
i'm doing AS Further Maths so if i've made assumptions that are over ambiguos or done anything wrong that i haven't realised yet then please correct me.

Prove that lnx ≠ x:
I thought the best way was proof by contradiction (alternative methods would be hugely appreciated) so;

If this is true then y = lnx would intersect the line y = x.
As y = e^x is a reflection of y = lnx and visa versa,
Then y = e^x would intersect y = lnx.

So when does:
lnx = e^x (apply e to both sides)
x = e^e^x (differentiate)
1 = e^x X e^e^x (ln both sides)
0 = ln(e^x X e^e^x)
0 = x + e^x
e^x = -x (Which is a contradiction)

So if lnx did = e^x then we would have a contradiction thus lnx ≠ e^x and thus there are no values of x such that lnx = x. End of Proof...I hope

[zaza]

Welcome to the Forums


Sorry that the first comment is negative, but I don't like your method. Particularly, I don't really like the step where you go to saying that the line y= e^x must intersect the line y = ln x.

A better way to do it would be the following.

1) The line y=ln x is meaningless for x <=0
2) The line y=x is positive for all x>0
3) At x<1, lnx is negative and hence <x
4) The line y=x has gradient = 1 for all x
5) At x=1, the line y=x has value 1, whereas the line y=ln x has value 0.
6) If we can then show that at no subsequent point does the gradient of y=lnx exceed 1, then we must conclude that the graph of lnx is never steeper than the graph of y=x and hence they can never cross.
7) The gradient of ln x can be obtained from its Taylor series.
8) Look up the Taylor expansion of ln x about x = a, you'll find it runs along the lines of :

f'(x) = 1/a - ((x-a)/a^2) + ((x-a)^2/a^3) - ...

Hence about any x=a, the gradient of y=ln x is 1/a = 1/x. This is a surprising, but well known fact.

So, in summary, for x<=1, y=x is positive whereas y=lnx <=0, hence the lines do not cross.
At x=1, y=x has a greater value than y=ln x.
For x>1, y=x has a gradient of 1, whereas y=lnx has a gradient of 1/x and hence for all x>1 the graph of y=ln x is never steeper than the graph of y=x, so the two lines do not cross for x>1.

Hence the two lines never cross.



zaza

[Andrew Roberts]

Yeah thats nice,
I was rather abstractly using the fact that e and ln are inverse functions to show that lnx ≠ x rather than relating it to y = x more. Algebraically i feel my method is concise and correct but ofcourse yourse is less abstract and far more useful.
Thanks for the "Taylor expansion of ln x about x = a:
f'(x) = 1/a - ((x-a)/a^2) + ((x-a)^2/a^3) - ...
Hence about any x=a, the gradient of y=ln x is 1/a = 1/x"
If i had know that then i would have used your method.
Thanks for that,
Andy

[Andrew Roberts]

Another question:
How do you go about working the inverse of a 3x3 matrix,
and knowing this is it obvious how to go onto find the inverse of a 4x4 matrix and so on...
I'm planning on using excel to help solve simultaneous equations containing more than 2 unknowns,
Many thanks
Andy

[zaza]

Here is a link to matrix inversion. Also look up the links therein to Gauss-Jordan elimination, LU decomposition etc, for an idea of how to proceed further with it. Also try a search on this website for Gauss-Jordan.



BTW, I do see what you were trying to accomplish with the inversion of ln x to e^x and the intersection of the two. You are trying to say that because one is the inverse of the other, that if one crosses the line y=x, then the other must do so at the same point and hence they must intersect each other. However, you need to bear in mind that this only applies because one is the reflection of the other in the line y=x, not because they are inverse (at least, not in the traditional sense of one-to-one or one-to many functions). It wouldn't work for any other line than y=x, unless you found the equivalent reflected function.


zaza