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Jinx
03-07-2006, 09:43 PM
Isn't it true that log(10^x) = x, right? If correct, then they are effectively mathematical opposites, right? And, if that is true, then:

So, shouldn't it be true that the graph of an exponential function would be observed to be an increasing function while the graph of a log function should be observed to be a decaying function, right? But, when looking at a log table, don't the values of log(x) increase as the value of (x) increases?

So, the right answer is that they are both increasing functions, only one is much steeper than the other, is that correct? I guess I never stopped to think about it before, but it seems a little contraintuitive to me. - Jinx

MrJackboots
03-07-2006, 09:49 PM
They're functional inverses, is all, which is to say that f(g(x))=x and g(f(x))=x. Doing one upon the other of a variable coughs the variable right back out. You're not multiplying one by the other, or adding them; I'm not entirely sure where the counterintuitive part comes in. I think you're overthinking it.

The same thing happens with F(x)=x-3 and G(x)=x+3, if you think about it:
F(G(x))=((X+3)-3)=x=((X-3)+3)=G(F(x))

Both of these increase as well, with the same slope even.

nivlac
03-07-2006, 09:57 PM
Isn't it true that log(10^x) = x, right? If correct, then they are effectively mathematical opposites, right? And, if that is true, then:

So, shouldn't it be true that the graph of an exponential function would be observed to be an increasing function while the graph of a log function should be observed to be a decaying function, right? But, when looking at a log table, don't the values of log(x) increase as the value of (x) increases?

So, the right answer is that they are both increasing functions, only one is much steeper than the other, is that correct? I guess I never stopped to think about it before, but it seems a little contraintuitive to me. - Jinx
The exponential and log functions are not mathematical "opposites." One function is the inverse function of the other. Just like the square and square root functions. One function "undoes" the other. There's no reason why both functions can't be increasing functions. In fact, if you're looking at the graphs of inverse functions, one is the mirror image of the other when reflected across the line y = x. And yes, for a given value of x>0 and a>1, the slope of y=ax is steeper than y=logax.

cerberus
03-07-2006, 11:07 PM
Functional inverses have a peculiar graphical symmetry, across the line y=x....

If we write the range values under the function f as {f(x)|x in the domain of F}, then the pairs (x,f(x)) form the graph of the function f. The inverse g of f works in the opposite direction, and the graph of g is obtained by transposing the the graph points of f: (x,y) from the graph of f is the point (y,x) of the graph of g, the inverse function of f. Functions whose inverses are also functions are said to be one-to-one.

The power laws, y=b^x, where b>0, have inverses, known as log-base-b.

log-base-b(y) = x if and only if b^x = y. If you were to graph the functions y=f(x)=b^x and y=g(x)=log-base-b(x) together with the line y=x, the graphs f and g are found to be symmetric across y=x. That is, if the point (z,w) is on the graph of f, then the point (w,z) is on the graph of g.

TJdude825
03-08-2006, 01:20 AM
Graph of y=ex

http://mathworld.wolfram.com/ExponentialFunction.html

Graph of y=ln x=logex

http://mathworld.wolfram.com/NaturalLogarithm.html

Now just imagine them on the same graph, and draw a diagonal line through the origin going in / direction. Now you've got a visual display of the inverse relationship described above.

Actually, now that I think about it, I'm pretty sure that if f(x) is always increasing for all x, its inverse will always increase as well. I can vaguely comprehend a way to prove this (if indeed it is true) using calculus, but there's got to be a non-calculus way also, I bet. Hm...

cerberus
03-08-2006, 02:03 AM
There may be an argument via first derivatives....

g(f(x)) = x
d/dx{g(f(x))} = 1
g'(f(x))f'(X) = 1

g'(f(x))= 1/f'(x) for all x in D(f) and f'(x) ne 0.

If we posit that f'(x) > 0 for all x in D(f), then g'(f(x)) > 0 for all x in D(f), hence g'(y) > 0 for all y in R(f). Increasing differentiable functions have positive first derivatives.

From a geometric point of view, the monotonically increasing bit, added to differentiabity forces each tangent line to y=f(x) to have a positive slope.

To wit, for each x in D(f), the tangent slope for the tangent line at (x,f(x)) has positive slope f'(x), and the formula is given as:

y = x*f'(x0)+b
f(x0)=x0*f'(x0)+b
b={f(x0)-x0*f'(x0)}
So, the tangent line to y=f(x) at x0 is given as y=x*f'(x0) + {f(x0)-x0*f'(x0)}.

What happens to a line when we map it across the line y=x?

y = mx + b becomes
x = ym + b becomes
x-b = ym becomes
y= (1/m)x - (b/m)...

Provided that m>0, the flipped slope 1/m > 0.

Sooooooooooo....the functional inverse of a differentiable function is increasing whenever the original function is itself increasing.

Malacandra
03-08-2006, 05:30 AM
Isn't it true that log(10^x) = x, right? If correct, then they are effectively mathematical opposites, right? And, if that is true, then:

So, shouldn't it be true that the graph of an exponential function would be observed to be an increasing function while the graph of a log function should be observed to be a decaying function, right? But, when looking at a log table, don't the values of log(x) increase as the value of (x) increases?

So, the right answer is that they are both increasing functions, only one is much steeper than the other, is that correct? I guess I never stopped to think about it before, but it seems a little contraintuitive to me. - Jinx

logy x increases as x increases, and yx increases as x increases, but the latter has a rapidly increasing gradient and the former has one that steadily decreases. That might be the "increase/decay" thing you're looking for. F'rinstance, if y=10, log x increases from 0 to 2 (two places) as x increases from 1 to 100 - but to get log x to increase two more places, you have to increase x all the way up to 10,000. And so on. Diminishing returns.

Thudlow Boink
03-08-2006, 07:18 AM
The exponential and log functions are not mathematical "opposites." One function is the inverse function of the other. Just like the square and square root functions. One function "undoes" the other. There's no reason why both functions can't be increasing functions.In fact (correct me if I'm wrong), the inverse of an increasing function has to be increasing as well: If bigger x's correspond to bigger y's, then bigger y's correspond to bigger x's. (in, for example, y = 10x)

chrisk
03-08-2006, 07:37 AM
In fact (correct me if I'm wrong), the inverse of an increasing function has to be increasing as well: If bigger x's correspond to bigger y's, then bigger y's correspond to bigger x's. (in, for example, y = 10x)

Well, let's try with a few simpler inverse pairs:

f(x) = x + 1, g(x) = x - 1

f(x) = x * 2, g(x) = x / 2

well, all of these are increasing functions, but I think they help demonstrate the point.

ultrafilter
03-08-2006, 08:50 AM
I can vaguely comprehend a way to prove this (if indeed it is true) using calculus, but there's got to be a non-calculus way also, I bet. Hm...

Proof without calculus:

First, any invertible function from R to R is either strictly increasing or strictly decreasing. If not, the intermediate value theorem can be used to show that the function is not one-to-one. It should also be obvious that both f and f-1 are invertible.

Let x1 < x2. For the moment, let's assume that f is strictly increasing, so f(x1) < f(x2). If f-1 is decreasing, then f-1(f(x1)) > f-1(f(x2)). In other words, x1 > x2. No dice.

Clearly, a very similar argument holds when f is strictly decreasing.

Jamaika a jamaikaiaké
03-08-2006, 04:41 PM
Proof without calculus:

... the intermediate value theorem ....

Eh? I beleive I taught the IVT to calc students last semester. Why? Because it involves continuity. Specifically, you need to assume the function is continuous in order to use the IVT.

Here is an invertible function which is neither strictly increasing nor strictly decreasing (t'ain't continuous neither)

f(x)= {1 if x=0; 0 if x=1; x otherwise}

OTOH, an increasing, invertible function from R to R must be continuous.

But we don't even need continuity: Assume f is increasing and invertible. If x<y and f-1(x)>f-1(y) then by increasingness of f, x=f(f-1(x))>f(f-1(y))=y, a contradiction.
(Keeping in mind, of course, that invertibility means that x /= y implies f(x) /= f(y) )