Example 25: Is $x=e^{\log x}$ for all real $x$⁠?

Basics you need

Solution

1. Where does each side make sense?

   The left side, $x$⁠, makes sense for every real $x$⁠. The right side, $e^{\log x}$⁠, needs $\log x$ first, and $\log x$ is defined only when $x > 0$⁠. So the right side exists exactly for

   $$x \in (0, \infty).$$
   An equation can only be true at an $x$ where both sides have a value. Before asking whether two expressions are equal, always ask where each one is defined. The book's observation (1) after Definition 4 says it plainly: there is no meaningful logarithm of a non-positive number.
   Domain of $\log$ is $\R^+$ (Section 5.4, observation (1) about the log function). Basics: exponentials and logarithms

2. For $x \le 0$ the statement fails

   Take $x = -2$⁠. Then $\log(-2)$ would have to be a number $t$ with $e^t = -2$⁠. But $e^t > 0$ for every real $t$⁠, so there is no such $t$⁠. The expression $e^{\log(-2)}$ has no value, and "$-2 = e^{\log(-2)}$" is not a true statement. The same happens for $x = 0$ (no $t$ gives $e^t = 0$⁠) and for every negative $x$⁠.

   One counterexample is enough to show that a "for all $x$" statement is false. We have in fact found that it fails for every $x \le 0$⁠, which tells us exactly where the trouble is.
   Range of the exponential function is the set of positive reals (Section 5.4, salient feature (2)). Basics: exponentials and logarithms

3. For $x > 0$ the statement is true

   Let $x > 0$ and put $t = \log x$⁠. By the definition of the logarithm,

   $$t = \log x \quad\text{means}\quad e^t = x.$$

   Replacing $t$ by $\log x$⁠: $e^{\log x} = x$⁠. So the equation holds for every positive $x$⁠.

   This is the most direct proof: the logarithm was defined as the exponent that $e$ needs to produce $x$⁠. Raising $e$ to that exponent therefore produces $x$⁠. The functions $e^x$ and $\log x$ are inverses of each other, and $e^{\log x} = x$ is "undo after do".
   Definition 4 (Section 5.4) with $b = e$⁠. Lesson: exponential and logarithmic functions

4. The book's argument, step by step

   The book argues differently. For $x > 0$ let $y = e^{\log x}$⁠. Then $y > 0$ (every power of $e$ is positive), so we may take its logarithm:

   $$\begin{aligned} \log y &= \log\left(e^{\log x}\right) \\ &= \log x\cdot\log e \\ &= \log x. \end{aligned}$$

   Since $\log$ is ever increasing, two different positive numbers have different logarithms, so $\log y = \log x$ forces $y = x$⁠.

   Each line needs a justification. Taking $\log y$ needs $y > 0$⁠. The second line uses $\log p^n = n\log p$ with $p = e$ and $n = \log x$ (the book proves it for positive integers $n$ and states, without proof, that it holds for every real $n$⁠). The third uses $\log e = 1$⁠, which is Definition 4 with base $e$⁠: $e^1 = e$⁠. The last step needs $\log$ to be one-one, which follows from it being ever increasing (observation (4)). Both arguments reach the same conclusion.
   $\log_b p^n = n\log_b p$ and the observations about $\log$ (Section 5.4). Basics: exponentials and logarithms

5. Conclude

   $x = e^{\log x}$ is not true for all real $x$⁠. It is true exactly for $x > 0$⁠, and meaningless for $x \le 0$⁠.

   Two functions are equal only if they have the same domain and agree at every point. $x$ and $e^{\log x}$ agree on $(0,\infty)$ but have different domains ($\R$ and $(0,\infty)$⁠). This matters later: logarithmic differentiation (Section 5.5) writes $u^v = e^{v\log u}$⁠, which is only allowed where $u > 0$⁠.
   Equality of functions: same domain and same value at each point (Chapter 1). Basics: exponentials and logarithms

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