1. ## 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.

2. ## Re: Divinding exponents

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.

3. ## Re: Divinding exponents

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

4. ## Re: Divinding 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.

5. ## Re: Divinding exponents

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)

6. ## Re: Divinding exponents

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

$\\x^3 = 0.216\\x^3 = \sqrt[3]{0.216} \\x^3 = (0.6)^3\\x = 0.6$

7. ## Re: Divinding exponents

And also: $a^x=a^y \Rightarrow x=y$.

Your solution is right, but your second step is incorrect, because if you say: $x^3=\sqrt[3]{0,216}$ that means $x^3=0,6$.
It has to be.
$x^3=0,216$
$\Leftrightarrow x=\sqrt[3]{0,216}$
$\Leftrightarrow x=0,6$.

8. ## Re: Divinding exponents

$\\x^3 = 0.216\\\sqrt[3]{x^3} = \sqrt[3]{0.216}\\ x = 0.6$

And like I said, is teh following true?

$x^a = 3^a, x = 3$?

9. ## Re: Divinding exponents

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. "

10. ## Re: Divinding exponents

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 $a\ne 0$ and $x, y > 0$. In fact, this follows from the previous statement. Indeed, $x^a=e^{a\ln x}$, so if $x^a=y^a$, then $e^{a\ln x}=e^{a\ln y}$, so $a\ln x=a\ln y$ and thus $\ln x=\ln y$ if $a\ne 0$. Now, ln is also a one-to-one function, so x = y.

11. ## Re: Divinding exponents

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