Q.Is it true that $x = e^{\log x}$ for all real $x$?

The key idea is the Inverse Function Relationship between the exponential function $e^x$ and the natural logarithm $\log x$ (where $\log$ denotes $\log_e$).

• The identity $e^{\log x} = x$ holds only when $\log x$ is defined, which requires $x > 0$ (the domain of $\log x$).
• For $x \leq 0$, $\log x$ is not a real number, so the expression $e^{\log x}$ is undefined in the real number system.
• Therefore, the statement is false for all real $x$ — it is true only for $x > 0$.

The core of this question lies in understanding the domain of the logarithmic function. In real analysis, $\log x$ (usually meaning the natural logarithm, $\log x$) is defined only for $x > 0$. This is not a technicality — it's a fundamental restriction because the exponential function $e^y$ is always positive, so its inverse can only accept positive inputs.

If you try to plug $x = 0$ or $x = -5$ into $\log x$, you get an undefined expression in the real number system. The equation $x = e^{\log x}$ therefore cannot even be considered for those values — it's like asking whether a square circle is round.

Let's walk through the reasoning step by step.

1. Recall the definition of the natural logarithm. The function $\log x$ (or $\log x$) is defined as the inverse of the exponential function $e^y$. That is:

For this to make sense, $x$ must be the output of $e^y$. Since $e^y > 0$ for every real $y$, the input $x$ to $\log x$ must be strictly positive: $x > 0$.

2. Check the identity on its natural domain. For any $x > 0$, the composition works perfectly:

This is the defining property of inverse functions — applying $e^y$ after $\log x$ returns the original $x$. So for all positive real numbers, the statement is true.

3. Test the boundary: $x = 0$.

   $\log 0$ is undefined (the limit as $x \to 0^+$ is $-\infty$, but it's not a real number). Therefore $e^{\log 0}$ is meaningless. The statement fails.

4. Test negative values: $x < 0$.

   $\log x$ for $x < 0$ is not defined in the real numbers (it exists in the complex plane, but that's a different story). So again, the expression $e^{\log x}$ is undefined. The statement fails.

5. Consider the converse: $x = \log(e^x)$.

   This is a different identity. $\log(e^x) = x$ holds for all real $x$, because $e^x$ is always positive and thus always in the domain of $\log$. But the original question asks about $e^{\log x}$, not $\log(e^x)$. These are not the same — the order of composition matters.

Method: Checking Whether an Inverse-Function Identity Holds Unconditionally

This is a reasoning method (no algebraic computation needed) for deciding whether a claimed identity like $f(f^{-1}(x)) = x$ or $f^{-1}(f(x)) = x$ is actually true for every real $x$, or only on a restricted domain.

Steps

Step 1: Identify the two functions involved and which is applied first

In $e^{\log x}$, the inner function is $\log x$ and the outer function is $e^{(\cdot)}$. Note the order — this is $f(f^{-1}(x))$ with $f(x) = e^x$, $f^{-1}(x) = \log x$.

Step 2: Recall the domain of the inner function, not just the outer one

Even if the outer function ($e^{(\cdot)}$) accepts every real number, the composite expression is only defined where the inner function is defined. Here, $\log x$ requires $x > 0$.

Step 3: State the identity only on the domain where it is actually defined

An identity between $f$ and its inverse can only be claimed on the domain where the composite expression makes sense in the first place — never on the outer function's full domain if that's wider than the inner function's domain.

Step 4: Conclude honestly about "for all real $x$" claims

If the required domain (Step 2) is narrower than "all real $x$," the statement as posed is false — even though it is true on the restricted domain.

Applying to this type of problem: the same check applies to any inverse-pair identity — e.g. $\sin^{-1}(\sin x) = x$ is also NOT true for all real $x$, only for $x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, for exactly the same reason: always check the domain of whichever function is applied first (innermost).

Common Mistakes

Mistake 1: Assuming $f(f^{-1}(x)) = x$ holds for every real $x$ just because $f$ and $f^{-1}$ are inverses

Why it's wrong: An inverse pair only "undoes" each other on the domain where the composition is actually defined — here, $e^{\log x}$ requires $\log x$ to exist first, which restricts $x$ to positive numbers, no matter how the outer exponential behaves. Correct approach: always check the domain of the innermost function applied (here $\log x$, needing $x>0$) before claiming an identity holds "for all $x$."

Mistake 2: Confusing $e^{\log x} = x$ with the different identity $\log(e^x) = x$

Why it's wrong: These look similar but are not the same statement — the order of composition matters. $\log(e^x)=x$ is true for all real $x$ (since $e^x$ is always positive, so it's always inside $\log$'s domain), while $e^{\log x}=x$ is true only for $x>0$. Correct approach: check which function is applied first in the given expression, and use that function's domain, not the other one's.