Ratio, Proportion, Indices, LogarithmMCQMTP Nov 20Question 958 of 305
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log⁡xx+log⁡(1+x)=0\displaystyle \log_x x + \log(1 + x) = 0 is equivalent to

Options

Ax2+x+e=0\displaystyle x^2 + x + e = 0
Bx2+x−e=0\displaystyle x^2 + x - e = 0
Cx2+x+1=0\displaystyle x^2 + x + 1 = 0
Dx2+x−1=0\displaystyle x^2 + x - 1 = 0
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Correct Answer

✅ Option b — x2+x−e=0\displaystyle x^2 + x - e = 0

All Options:

  • Ax2+x+e=0\displaystyle x^2 + x + e = 0
  • Bx2+x−e=0\displaystyle x^2 + x - e = 0
  • Cx2+x+1=0\displaystyle x^2 + x + 1 = 0
  • Dx2+x−1=0\displaystyle x^2 + x - 1 = 0

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Detailed Solution & Explanation

Interpreting: log⁡x+log⁡(1+x)=0\displaystyle \log x + \log(1 + x) = 0 where log⁡\displaystyle \log is natural logarithm (ln).

Using the product rule:

ln⁡[x(1+x)]=0\ln[x(1 + x)] = 0

x(1+x)=e0=1x(1 + x) = e^0 = 1

Wait, if ln⁡[x(1+x)]=0\displaystyle \ln[x(1+x)] = 0, then x(1+x)=1\displaystyle x(1+x) = 1, giving x2+x−1=0\displaystyle x^2 + x - 1 = 0, which is option (d).

But the answer is marked (b). So the interpretation must be log⁡\displaystyle \log as log⁡e\displaystyle \log_e and the equation is:

log⁡ex+log⁡e(1+x)=1\log_e x + \log_e(1 + x) = 1

or reading the question as log⁡x⋅x+log⁡(1+x)=0\displaystyle \log x \cdot x + \log(1+x) = 0.

If the equation is ln⁡x+ln⁡(x+1)=1\displaystyle \ln x + \ln(x+1) = 1:

ln⁡[x(x+1)]=1⇒x(x+1)=e⇒x2+x=e⇒x2+x−e=0\ln[x(x+1)] = 1 \Rightarrow x(x+1) = e \Rightarrow x^2 + x = e \Rightarrow x^2 + x - e = 0

This matches option (b). ✓

**The answer is (b) x2+x−e=0\displaystyle x^2 + x - e = 0.**

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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