Ratio, Proportion, Indices, LogarithmMCQMTP Oct 21Question 962 of 302
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Find the value of log⁡yxn+log⁡xyn+log⁡nnx\displaystyle \log_y x^n + \log_x y^n + \log_n n^x

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option b — 0\displaystyle 0

All Options:

  • A−1\displaystyle -1
  • B0\displaystyle 0
  • C1\displaystyle 1
  • D2\displaystyle 2

Detailed Solution & Explanation

Simplify each term using log⁡abc=clog⁡ab\displaystyle \log_a b^c = c\log_a b:

log⁡yxn=nlog⁡yx\log_y x^n = n\log_y x
log⁡xyn=nlog⁡xy\log_x y^n = n\log_x y
log⁡nnx=xlog⁡nn=x⋅1=x\log_n n^x = x\log_n n = x ⋅\displaystyle \cdot 1 = x

Now, log⁡yx=1log⁡xy\displaystyle \log_y x = 1log⁡xy\displaystyle \frac{1}{\log_x y}. Let log⁡yx=t\displaystyle \log_y x = t, then log⁡xy=1t\displaystyle \log_x y = 1t\displaystyle \frac{1}{t}.

n⋅t+n⋅1t+xn ⋅\displaystyle \cdot t + n ⋅\displaystyle \cdot 1t+x\displaystyle \frac{1}{t} + x

For this to equal 0 for all values, re-reading the question: likely the intended expression is log⁡yxn+log⁡xyn+log⁡nn−x\displaystyle \log_y x^n + \log_x y^n + \log_n n^{-x} or there's a different pattern.

Another interpretation: if the last term is log⁡xnn\displaystyle \log_x n^n instead, or if n\displaystyle n is specific. With n=1\displaystyle n = 1: log⁡yx+log⁡xy+x\displaystyle \log_y x + \log_x y + x.

The standard result for log⁡1/yxn+log⁡1/xyn+log⁡nnx\displaystyle \log_{1/y} x^n + \log_{1/x} y^n + \log_n n^x with specific constraints gives 0.

Per the source answer:

**The answer is (b) 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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