Q.Evaluate: ∫e4logx−e3logxe6logx−e5logxdx
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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: …
The key idea is to simplify the exponentials using enlogx=xn, then reduce the rational expression.
First, rewrite each term:
e6logx=x6,e5logx=x5,e4logx=x4,e3logx=x3.
So the integral becomes:
∫x4−x3x6−x5dx=∫x3(x−1)x5(x−1)dx. …
Since eklogx=xk, the integrand simplifies to x2, so the integral is 3x3+C.
Rewrite the exponentials. Using eklogx=xk:
∫e4logx−e3logxe6logx−e5logxdx=∫x4−x3x6−x5dx.
Simplify the rational function. Factor numerator and denominator: …
Method: Simplify enlogx before integrating
Use this whenever an integrand hides powers of x inside exponentials of logarithms. The integral looks intimidating but collapses to an elementary one after one simplification.
Steps
Step 1: Apply enlogx=xn.
Because log and exp are inverses, enlogx=(elogx)n=xn. Rewrite every such term.
Step 2: Factor numerator and denominator.
After converting, you get a rational function in x. Factor out common powers, e.g. x4−x3x6−x5=x3(x−1)x5(x−1).
Step 3: Cancel common factors. …
Common Mistakes
Mistake 1: Not converting enlogx to xn.
Why it's wrong: leaving the exponentials in place makes the integral look non-elementary, and students give up or misuse exponential rules. Correct approach: apply enlogx=xn first.
Mistake 2: Cancelling incorrectly across the fraction. …
- COMEDK 2023Set 2023-M1 markMCQQ.The value of alogbc−clogba, where a,b,c>0 but a,b,c=1, is (A) a (B) b (C) c (D) 0
›Reveal solutionSolution
alogbc and clogba are equal, so their difference is zero.
Take logarithm (base b) of the first term:
logb(alogbc)=(logbc)(logba).
For the second term:
logb(clogba)=(logba)(logbc). …
- COMEDK 2022Set 20221 markMCQQ.The value of 3log45−5log43 is (A) 0 (B) 1 (C) 2 (D) 4
›Reveal solutionSolution
3^(log_4 5) - 5^(log_4 3) = 0
Concept: The identity a^(log_c b) = b^(log_c a).
Proof: take log to base c of the left side: log_c(a^(log_c b)) = (log_c b)(log_c a), which is symmetric in a and b, so it equals log_c(b^(log_c a)). Since log is injective, a^(log_c b) = b^(log_c a). …
- KCET 2021Set A-11 markMCQQ.Consider the following statements: Statement 1: If y=log10x+logex then dxdy=xlog10e+x1 Statement 2: dxd(log10x)=log10logx and dxd(logex)=logelogx (A) Statement 1 is true; statement 2 is false (B) Statement 1 is false; statement 2 is true (C) Both statements 1 and 2 are true (D) Both statements 1 and 2 are false
›Reveal solutionSolution
Differentiate using the change-of-base rule logax=lnalnx; Statement 1 checks out, while Statement 2 quotes the function itself instead of its derivative.
Step 1 — The concept: derivative of a logarithm to any base.
Only the natural logarithm has the clean derivative dxd(lnx)=x1. For any other base a we first convert with the change-of-base identity
logax=lnalnx
Here lna is a constant, so it just rides along:
dxd(logax)=lna1⋅x1=xlogae,
using lna1=logae.
Step 2 — Test Statement 1.
With y=log10x+logex,
dxdy=dxd(log10x)+dxd(logex)=xlog10e+x1.
This is exactly what Statement 1 claims. Statement 1 is TRUE.
Step 3 — Test Statement 2.
Statement 2 asserts
dxd(log10x)=log10logx,dxd(logex)=logelogx. …
- COMEDK 2021Set 20211 markMCQQ.83log85 is equal to (A) log825 (B) 120 (C) 125 (D) log815
›Reveal solutionSolution
8^(3 log_8 5) = 8^(log_8 5^3) [bring the 3 inside as an exponent] = 8^(log_8 125) = 125 [since a^(log_a x) = x]
Concept: a^(log_a x) = x, and n log_a x = log_a (x^n).
8^(3 log_8 5)
= 8^(log_8 5^3) [bring the 3 inside as an exponent] …
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