Q.log∫01e100dx=
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Exponential Logarithmic Simplification
Exponential Logarithmic Simplification
You've probably seen expressions like elogx or log(ex) 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 ex and the natural logarithm logx 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, elogx=x.
- Start with a number, exponentiate it, then take the natural log: you also get back, log(ex)=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 functions ex and logx are inverses — they reverse each other's effect, just like x and x2 are inverses for x≥0.
The Precise Statement
elogx=xfor all x>0
log(ex)=xfor all real x
The first formula works only when x>0 because logx is only defined for positive inputs. The second works for any real x because ex is always positive.
A common mistake is to write elogx=x for x≤0. This is wrong — logx is undefined for x≤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: elogx=e5⟹x=e5.
- To solve ex=7, take the natural log: log(ex)=log7⟹x=log7.
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 elog(3x+1). The expression is defined only when 3x+1>0; if that holds, then: …
e100 is a constant, so ∫01e100dx=e100; then loge100=100. …
e100 is a constant, so ∫01e100dx=e100; then loge100=100.
Since e100 is a constant, ∫01e100dx=e100⋅(1−0)=e100. Taking the n …
- CBSE 2026Set ANNUAL1 markMCQQ.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
elog(anything) simplifies to that "anything" first, then integrate.
Concept: elogf(x)=f(x) since log and e(⋅) are inverse functions.
Working: …
- CBSE 2026Set ANNUAL1 markMCQQ.∫e5logxdx is equal to(a) 5x5+C(b) 6x6+C(c) 5x4+C(d) 6x5+C
›Reveal solutionSolution
e5logx=x5, so the integral is 6x6+C.
Step 1: e5logx=elogx5=x5 (for x>0).
…
- CBSE 2025Set E1 markMCQQ.∫e2logxdx=(a) e2logx+k(b) 2x2+k(c) 3x3+k(d) 3x3+k
›Reveal solutionSolution
e2logx=x2, and ∫x2dx=3x3+k.
Use alogx=logxa and elogy=y:
e2logx=elogx2=x2.
Therefore …
- CBSE 2025Set E1 markMCQQ.log∫01e100dx=(a) 100(b) 1001(c) 1(d) 101
›Reveal solutionSolution
e100 is a constant, so ∫01e100dx=e100; then loge100=100.
Since e100 is a constant, ∫01e100dx=e100⋅(1−0)=e100. Taking the n …
- CBSE 2025Set ANNUAL1 markMCQQ.∫ 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 2logx as a power of x using alogb=bloga, then integrate a simple power.
Using the identity alogb=bloga (both equal e(loga)(logb)), with a=2, b=x:
2logx=xlog2.
…
- CBSE 2020Set ANNUAL1 markQ.Solve: ∫a3logaxdx.
›Reveal solutionSolution
Simplify a3logax to a plain power of x first, using alogax=x, then integrate the power.
Step 1 — simplify the integrand.
Since alogax=x for any valid base a,
a3logax=alogax3=x3
Step 2 — integrate. …
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