∫ 2^(log x) dx equals – (i) 2^(log x + 1) / (log x – Class 12 Mathematics

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Q. \int 2^(log x) dx equals -

(i) 2^(log x + 1) / (log x + 1) + C

(ii) x^(log 2 + 1) / (log 2 + 1) + C

(iii) 2^(log x) / log 2 + C

(iv) 2^(log x) / 2 + C

Mizoram MbseMizoram Board of School Education HSSLC 2025MCQ· 1mImportance★★★★★

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Concept understanding — Exponential Logarithmic Simplification

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).

Note

The functions $e^x$ and $\log x$ are inverses — they reverse each other's effect, just like $\sqrt{x}$ and $x^2$ are inverses for $x \geq 0$ .

The Precise Statement

e^{\log x} = x \quad \text{for all } x > 0

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.

Watch out

A common mistake is to write $e^{\log x} = x$ for $x \leq 0$ . This is wrong — $\log x$ is undefined for $x \leq 0$ in the reals. Always check the domain.

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: …

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Rewrite $2^{\log x}$ as a power of $x$ using $a^{\log b}=b^{\log a}$ , then integrate a simple power.

Using the identity $a^{\log b}=b^{\log a}$ (both equal $e^{(\log a)(\log b)}$ ), with $a=2,\ b=x$ :

$2^{\log x}=x^{\log 2}$ .

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q.The value of \int e^{\log(\cos x)},dx will be:$
(a) -\sin x + c
(b) e^{\log(\sin x)} + c
(c) \sin x + c
(d) \sec x + c
›Reveal solutionSolution

$e^{\log(\text{anything})}$ simplifies to that "anything" first, then integrate.

Concept: $e^{\log f(x)} = f(x)$ since $\log$ and $e^{(\cdot)}$ are inverse functions.

Working: …

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CBSE 2026Set ANNUAL1 markMCQ
$\int e^{5\log x}\,dx$
(a) $\dfrac{x^{5}}{5}+C$
(b) $\dfrac{x^{6}}{6}+C$
(c) $5x^{4}+C$
(d) $6x^{5}+C$
›Reveal solutionSolution

$e^{5\log x}=x^5$ , so the integral is $\dfrac{x^6}{6}+C$ .

Step 1: $e^{5\log x}=e^{\log x^{5}}=x^{5}$ (for $x>0$ ).

…

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CBSE 2025Set E1 markMCQ
$\int e^{2\log x}\,dx =$
(a) $e^{2\log x} + k$
(b) $\frac{x^2}{2} + k$
(c) $\frac{x^3}{3} + k$
(d) $3x^3 + k$
›Reveal solutionSolution

$e^{2\log x} = x^2$ , and $\int x^2\,dx = \frac{x^3}{3}+k$ .

Use $a\log x = \log x^a$ and $e^{\log y} = y$ :

$e^{2\log x} = e^{\log x^2} = x^2.$

Therefore …

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CBSE 2025Set E1 markMCQ
$\log \int_{0}^{1} e^{100}\,dx =$
(a) $100$
(b) $\frac{1}{100}$
(c) $1$
(d) $101$
›Reveal solutionSolution

$e^{100}$ is a constant, so $\int_0^1 e^{100}dx=e^{100}$ ; then $\log e^{100}=100$ .

Since $e^{100}$ is a constant, $\int_0^1 e^{100}\,dx = e^{100}\cdot(1-0) = e^{100}.$ Taking the n …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. \int 2^(log x) dx equals -$
(i) 2^(log x + 1) / (log x + 1) + C
(ii) x^(log 2 + 1) / (log 2 + 1) + C
(iii) 2^(log x) / log 2 + C
(iv) 2^(log x) / 2 + C
›Reveal solutionSolution

Rewrite $2^{\log x}$ as a power of $x$ using $a^{\log b}=b^{\log a}$ , then integrate a simple power.

Using the identity $a^{\log b}=b^{\log a}$ (both equal $e^{(\log a)(\log b)}$ ), with $a=2,\ b=x$ :

$2^{\log x}=x^{\log 2}$ .

…

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CBSE 2020Set ANNUAL1 mark
$\displaystyle\int a^{3\log_a x}\,dx$
›Reveal solutionSolution

Simplify $a^{3\log_a x}$ to a plain power of $x$ first, using $a^{\log_a x}=x$ , then integrate the power.

Step 1 — simplify the integrand.

Since $a^{\log_a x}=x$ for any valid base $a$ ,

$a^{3\log_a x} = a^{\log_a x^3} = x^3$

Step 2 — integrate. …

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Q.∫ 2^(log x) dx equals –

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