Thread: Divinding exponents

Hey guys, for the following question how do I solve for x?

(2^3)^x = 2^6

I know that to raise a power to a power I need to multiply the exponents. In the above question I thought that maybe I can solve it by dividing the exponents, but my math book doesn't mention anything about dividing exponents.

here's how I tried to solve it:
(2^3)^x = 2^6
(2^3)^x = (2^3)^2
x = 2

I assumed that since both have the same bases, I can equate x to 2.

Your solution is correct.

I assumed that since both have the same bases, I can equate x to 2.

Yes, exponentiation is a one-to-one function: if a > 1 or 0 < a < 1, then a^x = a^y implies x = y for all x and y.

Thanks emakarov. I have another question: is it possible to divide exponents?

Originally Posted by darksoulzero

I have another question: is it possible to divide exponents?

What do you mean by this? The right question in mathematics is whether something is true, not if one can do something.

Originally Posted by darksoulzero

I have another question: is it possible to divide exponents?

If you mean a^x / a^y, then: a^(x-y)

Ok, I see thanks guys.

based on what emakarov said, if a>1, or 0<a<1, a^x = a^y, then x = y, is the following also true?

x^a = y^a, then x = y?

here's what i tried

And also: .

Your solution is right, but your second step is incorrect, because if you say: that means .
It has to be.


.

What I meant for the second step was this:



And like I said, is teh following true?

?

Yes, you were told that in the very first response:

emakarov said
"Yes, exponentiation is a one-to-one function: if a > 1 or 0 < a < 1, then a^x = a^y implies x = y for all x and y. "

Originally Posted by darksoulzero

based on what emakarov said, if a>1, or 0<a<1, a^x = a^y, then x = y, is the following also true?

x^a = y^a, then x = y?

Yes, if and . In fact, this follows from the previous statement. Indeed, , so if , then , so and thus if . Now, ln is also a one-to-one function, so x = y.

Ok, I see. thanks emakarov. And thanks to everyone else that helped.