1
$\begingroup$

I am not sure on how to derive the following statement concerning the reals (that I think should be true).

For every real $y > 0$, for every positive real $z \neq 1$, there is a $x \in \mathbb{R}$ such that $y=z^x$.

QUESTIONS:

  • Is this part of the definition of exponentiation, or it is a result that can be proven?

  • If it can be proven, how (maybe by using the Archimedean Property)?

Thank you for your time.

PS: Few days ago I asked almost the same question, but a user pointed out (rightly) that the question I really wanted to ask is the one here.

$\endgroup$
6
  • $\begingroup$ Actually, as always, there were typos in my question, and you pointed them. I really had in mind the exponential function, hence what you wrote. Hence, I will correct it, because I still don't see how to obtain the correct statement. $\endgroup$
    – Kolmin
    Jan 6, 2016 at 15:22
  • 2
    $\begingroup$ So you're just asking why the exponential functions $f_z : \mathbf{R}^+ \to \mathbf{R}^+$, given by $f_z (x) = z^x$ is surjective. You can show more by producing an inverse, namely $g_z (x) = \log _ z (x)$. $\endgroup$
    – Future
    Jan 6, 2016 at 15:28
  • 1
    $\begingroup$ The $x$ you want can be expressed as $\frac{\ln y}{\ln z}$. Exactly how one proves the result depends on the details of how one develops the theory of the exponential function and logarithm. $\endgroup$ Jan 6, 2016 at 15:33
  • $\begingroup$ Thanks a lot both for your feedbacks! $\endgroup$
    – Kolmin
    Jan 6, 2016 at 15:39
  • $\begingroup$ @T.S.L Shouldn't it be $f_z : \mathbb{R} \to \mathbb{R}*$? $\endgroup$
    – Kolmin
    Jan 6, 2016 at 15:40

3 Answers 3

1
$\begingroup$

Since the exponential function $\exp(t) = e^t$ strictly increases on $\mathbb{R}$, (Check this in your analysis textbook) its inverse, the logarithmic function $\log t$ on $\mathbb{R}^+$ is well-defined. If we take $x = \log y / \log z$, $ z^x = e^{\log y} = y$ clearly.

$\endgroup$
0
$\begingroup$

You can use the fact that $\mathbb{R}$ has the least upper bound property. Let $E$ be the set of all $t$ such that $t^n \lneq x$, then if we can show that $E$ is nonempty and bounded above, we are guaranteed the existence of $y =sup$ $E$ in $\mathbb{R}$. From there, if we assume that $ y^n \lneq x$ or $y^n \gneq x$ we should be able to obtain a contradiction (or else we would have produced a "gap" in the real line). The uniqueness part of the proof is much easier since if a number had two distinct real nth roots, then one would necessarily be larger than the other.

Also this is not a part of the definition of exponentiation. Note that we can take exponents in $\mathbb{Q}$ but the statement you asked about no longer holds. So this is indeed something that must be proven.

$\endgroup$
-1
$\begingroup$

$$y = a^{\log_a(y)} \to z=a, x=\log_a (y)$$

$\endgroup$

You must log in to answer this question.

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