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Logarithms

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Question : 42 of 85

Marks: +1, -0

The value of

\log_5\left(1+\;\frac{1}{5}\right)+\log_5\left(1+\;\frac{1}{6}\right)+\log_5\left(1+\;\frac{1}{7}\right)+\dots+\log_5\left(1+\;\frac{1}{624}\right)

is

5
4
3
2

Solution:

Concept:

Rewrite each term as

\log_5\left(\frac{n+1}{n}\right)

and use the product property of logarithms.

Explanation:

Each term is

\log_5\left(1+\frac{1}{k}\right) = \log_5\left(\frac{k+1}{k}\right)

for

k = 5, 6, 7, \dots, 624

.

The sum is

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

.

Using

\log a + \log b = \log(ab)

, the sum equals

\log_5\left(\frac{6}{5} \cdot \frac{7}{6} \cdots \frac{625}{624}\right)

.

Telescoping gives

\frac{625}{5} = 125

.

Thus, the sum is

\log_5 125 = \log_5 5^3 = 3

.

Answer:

The value is

3

.

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