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Logarithms

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Question : 43 of 85

Marks: +1, -0

If

x=1983!

, then the value of the expression

\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \frac{1}{\log_4 x} + \cdots + \frac{1}{\log_{1983} x}

is equal to

0
1
2
3

Solution:

Concept:

Use the property

\frac{1}{\log_k x} = \log_x k

.

Explanation:

Given

x = 1983!

.

The sum is

\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \cdots + \frac{1}{\log_{1983} x}

.

By the change of base formula,

\frac{1}{\log_k x} = \log_x k

.

Thus the sum becomes

\log_x 2 + \log_x 3 + \cdots + \log_x 1983

.

Using the logarithm addition property, this equals

\log_x (2 \cdot 3 \cdots 1983) = \log_x (1983!)

.

Since

x = 1983!

, we have

\log_x x = 1

.

Answer:

The value is 1, which corresponds to option B.

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