5
$\begingroup$

I'm trying to see why the below is true.

$$ x^{\log(a)} = a^{\log(x)} $$

Anyone here know why this is? Thank you.

$\endgroup$
3
  • $\begingroup$ Incidentally, you can use this to put a field structure on $\mathbb{R}_{\geq 0}$, where the "addition" operation is given by $x\cdot y$ with identity $1$, and the "multiplication" operation is given by $x^{\ln(y)}$ with identity $e$. This fact is what makes the "multiplication" commutative. $\endgroup$
    – 2'5 9'2
    Jan 31, 2015 at 1:08
  • $\begingroup$ @alex.jordan, you're almost right, but the field structure is on $\mathbb{R}_{>0}$. See my answer. $\endgroup$ Jan 31, 2015 at 2:25
  • $\begingroup$ @goblin Oh yeah, of course. Just writing too fast without thinking. $\endgroup$
    – 2'5 9'2
    Jan 31, 2015 at 4:20

4 Answers 4

15
$\begingroup$

$$\large x^{\log(a)} = (e^{\log(x)})^{\log(a)} = e^{\log(x)\log(a)} = a^{\log(x)}$$

This only works if $x$ and $a$ are both positive real numbers.

$\endgroup$
3
  • $\begingroup$ @PublicEnemy I just applied the definition of the logarithm (rewrote x in term of log) then applied $(x^a)^b=x^{ab}$ law, followed by commutativity, simplified other term... $\endgroup$ Jan 30, 2015 at 23:56
  • $\begingroup$ @sanjab, What if the base of $\log a$ is not $e?$ $\endgroup$ Jan 31, 2015 at 3:03
  • 1
    $\begingroup$ @labbhattacharjee then the calculation goes the same: $x^{\log_b(a)} = (b^{\log_b(x)})^{\log_b(a)} = b^{\log_b(x)\log_b(a)} = a^{\log_b(x)}$ $\endgroup$ Jan 31, 2015 at 15:38
5
$\begingroup$

take the $\log$ for both sides to get $$\log (x^{log(a)})=\log (a^{\log(x)})$$ $$\log (a){log(x)}=\log( a){\log(x)}$$ It is clear that the $a$,$x$ should be positive

$\endgroup$
3
$\begingroup$

This is essentially another way of saying what sanjab has already said, but in a way that gives it a bit more intellectual context. Its sort of the "deeper reason" why it works. So why does $p^{\log(q)} = q^{\log(p)}$? Well, because there's this commutative operation $$\otimes : \mathbb{R}_{>0} \times \mathbb{R}_{>0} \longrightarrow \mathbb{R}_{>0}$$ that I'm about to define, and as it turns out, we have the following.

$$p^{\log(q)} = p \otimes q$$ $$q^{\log(p)} = q \otimes p$$

So let me go ahead and explain the notation. Define $\otimes$ as follows.

$$p \otimes q = \exp(\log(p)\log(q))$$

Then $\otimes$ is associative and commutative, and it has an identity element, namely $e$, by which I mean that the following identity holds, for all $p \in \mathbb{R}_{>0}$.

$$e \otimes p = p \otimes e = p$$

Now define that $p^* = \exp(1/\log(p))$. This makes sense whenever $p \in \mathbb{R}_{>0}$ is distinct from $1$, in which case:

$$p^* \otimes p = p \otimes p^* = e$$

Furthermore, $\otimes$ distributives over multiplication:

$$p \otimes (qr) = (p \otimes q)(p \otimes r)$$

If this all seems like magic, don't worry, I'm about the take the magic away. There is a homomorphism of abelian groups

$$\exp : (\mathbb{R},+,0) \rightarrow (\mathbb{R}_{>0},\times,1).$$

In fact, this function is bijective, so we can push operations on $\mathbb{R}$ across $\exp$ to obtain operations on $\mathbb{R}_{>0}.$ Multiplication on $\mathbb{R}_{>0}$ is what we get for pushing $+_\mathbb{R}$ across $\exp$, and $\otimes$ is what we get for pushing $\times_\mathbb{R}$. So in fact, $\mathbb{R}_{>0}$ is completely isomorphic to $\mathbb{R}$.

Now given $p \in \mathbb{R}_{>0}$ and $x \in \mathbb{R}$, we can define $p^x \in \mathbb{R}_{>0}$ as follows:

$$p^x = p \otimes \exp x$$

Getting back to your question, we have:

$$p^{\log(q)} = p \otimes \exp (\log q) = p \otimes q$$ $$q^{\log(p)} = q \otimes \exp (\log p) = q \otimes p$$

$\endgroup$
2
$\begingroup$

Yes: the general definition of $A^B$ is $\exp(B\log(A))$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .