Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

#1 2005-12-26 07:41:40

se7en
Member

Offline

exponential equations

In a book I'm reading an exponential equation is defined as an equation containing terms of the form a^x (a to the power of x), where x is a real number, a is a real number, a > 0 and a is not equal to 1.

Now I understand why the definition excludes a < 0 (there are many values of x for which a^x is undefined if a < 0, for example if a = -2 and x = 1/2, a^x = (-2)^1/2 = sqrt(-2) which is undefined.)

But I can't understand why the definition excludes a = 0 and a = 1. The following reasons are given in the book (which I don't understand... reasons are given).

If a = 0 then a^x becomes 0^x. This is equal to the constant 0 for all x except x = 0. It is undefined for x = 0. Thus when a = 0, a^x reduces to a constant or is undefined.

Reason why I don't understand the above: Even if a = 0 in a^x = a^y, surely x still equals y.

If a = 1, then a^x = 1^x = 1 for all real values of x. In this case a^x reduces to the constant 1.

Reason why I don't understand the above: Again, even if a = 1 in a^x = a^y, surely x still equals y.

Now I have an idea of my own on why a = 0 and a = 1 are excluded in the definition and I would appreciate your input on the idea. My idea is that if a = 0 or a = 1 the equation is true for all values of x.

#2 2005-12-26 08:19:58

mathsyperson
Moderator

Offline

Re: exponential equations

Did you get that maths book as a Christmas present? wink

Anyway, the book is right. Just think about it.

0² = 0³, because they both equal 0, but 2 ≠ 3.
Similarly, 1² = 1³, because they both equal one, but 2 ≠ 3.

A more mathematical approach would be to try to solve an equation like that.

1^x = 1^y.
This means that x = log y ÷ log 1

But, as log 1 = 0, then you can't divide by it.

If you try to do a similar thing with a = 0, you hit the problem that log 0 is undefined.

Happy Christmas!
       ,--.
      () / `\
        /      \
   _ /______\_
  (__________)
    (/  @  @  \)
    (`._,()._,' )
     (  `-'`-'  )
       \        /
        \,,,,,,/


Why did the vector cross the road?
It wanted to be normal.

#3 2005-12-26 16:40:11

Ricky
Moderator

Offline

Re: exponential equations

Reason why I don't understand the above: Even if a = 0 in a^x = a^y, surely x still equals y.

Nope, not true.  0^4 = 0^5.

Reason why I don't understand the above: Again, even if a = 1 in a^x = a^y, surely x still equals y.

Yea, but can you really call this exponential?  If you do, then just about every single equation is exponential.  Example:

y = x^2 + 2.

Certainly doesn't seem exponential, does it?  But watch:

y = x^2 + 2 * (1^x)

Since for all values, 1^x = 1, these are the same equations.  Now would you call this exponential?  Nah, it makes no sense to.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Board footer

Powered by FluxBB