If x = log24 12 , y = log36 24 , z = log48 36 , then xyz + 1 =

Option c: 2yz. Solution: Using change of base formula: x = ln 12/ln 24, y = ln 24/ln 36, z = ln 36/ln 48 xyz = ln 12/ln 24 × ln 24/ln 36 × ln 36/ln 48 = ln 12/ln 48 xyz + 1 = ln 12/ln 48 + 1 = ln 12 + ln 48/ln 48 = ln(12 × 48)/ln 48 = ln 576/ln 48 Since 576 = 242 : ln 576 = 2ln 24 . xyz + 1 = 2ln 24/ln 48 Now check 2yz : 2yz = 2 × ln 24/ln 36 × ln 36/ln 48 = 2ln 24/ln 48 Therefore xyz + 1 = 2yz . ✓ The answer is (c) 2yz .

Ratio, Proportion, Indices, LogarithmMCQMTP June 22Question 964 of 305

All Questions

If $\displaystyle x = \log_{24} 12$ , $\displaystyle y = \log_{36} 24$ , $\displaystyle z = \log_{48} 36$ , then $\displaystyle xyz + 1 =$

Options

\displaystyle 2xy

\displaystyle 2xz

\displaystyle 2yz

\displaystyle 2

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Correct Answer

✅ Option c — $\displaystyle 2yz$

All Options:

A $\displaystyle 2xy$
B $\displaystyle 2xz$
C $\displaystyle 2yz$
D $\displaystyle 2$

Detailed Solution & Explanation

Using change of base formula:

x = \frac{\ln 12}{\ln 24}, \quad y = \frac{\ln 24}{\ln 36}, \quad z = \frac{\ln 36}{\ln 48}

xyz = \frac{\ln 12}{\ln 24} \cdot \frac{\ln 24}{\ln 36} \cdot \frac{\ln 36}{\ln 48} = \frac{\ln 12}{\ln 48}

xyz + 1 = \frac{\ln 12}{\ln 48} + 1 = \frac{\ln 12 + \ln 48}{\ln 48} = \frac{\ln(12 \times 48)}{\ln 48} = \frac{\ln 576}{\ln 48}

Since

\displaystyle 576 = 24^2

\displaystyle \ln 576 = 2\ln 24

xyz + 1 = \frac{2\ln 24}{\ln 48}

Now check

\displaystyle 2yz

2yz = 2 \cdot \frac{\ln 24}{\ln 36} \cdot \frac{\ln 36}{\ln 48} = \frac{2\ln 24}{\ln 48}

Therefore

\displaystyle xyz + 1 = 2yz

. ✓

**The answer is (c)

\displaystyle 2yz

.**

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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If x=log⁡2412\displaystyle x = \log_{24} 12x=log24​12, y=log⁡3624\displaystyle y = \log_{36} 24y=log36​24, z=log⁡4836\displaystyle z = \log_{48} 36z=log48​36, then xyz+1=\displaystyle xyz + 1 =xyz+1=