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**se7en****Member**- Registered: 2005-12-25
- Posts: 28

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.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Did you get that maths book as a Christmas present?

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!

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Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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..."

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