The value of log5 (1 + 1/5) + log5 (1 + 1/6) + ... + log5 (1 + 1/624)

Option b: 3. Solution: We are given the summation of logarithms: S = log5 (1 + 1/5) + log5 (1 + 1/6) + ... + log5 (1 + 1/624) Let us first simplify the fractions inside each logarithm: 1 + 1/5 = 6/5 1 + 1/6 = 7/6 1 + 1/7 = 8/7 ... 1 + 1/624 = 625/624 Substitute these simplified fractions back into the summation: S = log5 ( 6/5 ) + log5 ( 7/6 ) + log5 ( 8/7 ) + ... + log5 ( 625/624 ) Using the logarithmic product rule, logb(x) + logb(y) + logb(z) + ... = logb(x × y × z × ...) : S = log5 ( 6/5 × 7/6 × 8/7 × ... × 625/624 ) This is a telescoping product, where the numerator of each term cancels out with the denominator of the subsequent term: S = log5 ( 625/5 ) S = log5 (125) Expressing 125 as a power of 5 (since 125 = 53 ): S = log5 ( 53 ) = 3 log5 5 = 3 × 1 = 3 Hence, Option B is the correct answer.

Ratio, Proportion, Indices, LogarithmMCQPYQ June 19Question 932 of 305

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The value of $\displaystyle \log_5 \left(1 + \frac{1}{5}\right) + \log_5 \left(1 + \frac{1}{6}\right) + \dots + \log_5 \left(1 + \frac{1}{624}\right)$

Options

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Correct Answer

✅ Option b — 3

All Options:

Detailed Solution & Explanation

We are given the summation of logarithms:

S = \log_5 \left(1 + \frac{1}{5}\right) + \log_5 \left(1 + \frac{1}{6}\right) + \dots + \log_5 \left(1 + \frac{1}{624}\right)

Let us first simplify the fractions inside each logarithm:

1 + \frac{1}{5} = \frac{6}{5}

1 + \frac{1}{6} = \frac{7}{6}

1 + \frac{1}{7} = \frac{8}{7}

\dots

1 + \frac{1}{624} = \frac{625}{624}

Substitute these simplified fractions back into the summation:

S = \log_5 \left( \frac{6}{5} \right) + \log_5 \left( \frac{7}{6} \right) + \log_5 \left( \frac{8}{7} \right) + \dots + \log_5 \left( \frac{625}{624} \right)

Using the logarithmic product rule,

\displaystyle \log_b(x) + \log_b(y) + \log_b(z) + \dots = \log_b(x \times y \times z \times \dots)

S = \log_5 \left( \frac{6}{5} \times \frac{7}{6} \times \frac{8}{7} \times \dots \times \frac{625}{624} \right)

This is a telescoping product, where the numerator of each term cancels out with the denominator of the subsequent term:

S = \log_5 \left( \frac{625}{5} \right)

S = \log_5 (125)

Expressing

\displaystyle 125

as a power of

\displaystyle 5

(since

\displaystyle 125 = 5^3

S = \log_5 \left( 5^3 \right) = 3 \log_5 5 = 3 \times 1 = 3

Hence, **Option B** is the correct answer.

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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The value of log⁡5(1+15)+log⁡5(1+16)+⋯+log⁡5(1+1624)\displaystyle \log_5 \left(1 + \frac{1}{5}\right) + \log_5 \left(1 + \frac{1}{6}\right) + \dots + \log_5 \left(1 + \frac{1}{624}\right)log5​(1+51​)+log5​(1+61​)+⋯+log5​(1+6241​)