Q.$\int e^{2\log x}\,dx =$

Exponential Logarithmic Simplification

You've probably seen expressions like $e^{\log x}$ or $\log(e^x)$ and wondered whether they just cancel out. The short answer is yes — but only under the right conditions. This is what we call exponential logarithmic simplification.

The Intuition

Think of the exponential function $e^x$ and the natural logarithm $\log x$ as inverse operations — they "undo" each other.

• Start with a number, take its natural log, then exponentiate the result: you get back where you started, $e^{\log x} = x$.
• Start with a number, exponentiate it, then take the natural log: you also get back, $\log(e^x) = x$.

This is exactly like how adding 5 and subtracting 5 cancel out, or how squaring and taking the square root undo each other (for non-negative numbers).

The Precise Statement

The first formula works only when $x > 0$ because $\log x$ is only defined for positive inputs. The second works for any real $x$ because $e^x$ is always positive.

Why This Matters

This simplification lets you solve equations that mix exponentials and logs:

• To solve $\log(x) = 5$, exponentiate both sides: $e^{\log x} = e^5 \implies x = e^5$.
• To solve $e^x = 7$, take the natural log: $\log(e^x) = \log 7 \implies x = \log 7$.

Without this rule you'd be stuck; with it, you can "peel away" the exponential or the log to isolate the variable.

A Quick Example

Simplify $e^{\log(3x+1)}$. The expression is defined only when $3x+1 > 0$; if that holds, then: …