Test Index

Logarithms

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Question : 26 of 85

Marks: +1, -0

The value of

\log_{3} 2 \;\log_{4} 3\; \log_{5} 4 . . . \log_{15} 14\; \log_{16}

$\; \frac{1}{3}$
$\; \frac{1}{2}$
$\; \frac{1}{5}$
$\; \frac{1}{4}$

Solution:

Concept:

Use change-of-base formula to telescope the product.

Explanation:

Write each term as

\log_a b = \frac{\ln b}{\ln a}

.

Then the product becomes:

\frac{\ln 2}{\ln 3} \cdot \frac{\ln 3}{\ln 4} \cdot \frac{\ln 4}{\ln 5} \cdots \frac{\ln 14}{\ln 15} \cdot \frac{\ln 15}{\ln 16}

.

All intermediate logarithms cancel, leaving

\frac{\ln 2}{\ln 16}

.

Now

\frac{\ln 2}{\ln 16} = \log_{16} 2

.

Since

16^{1/4} = 2

, we have

\log_{16} 2 = \frac{1}{4}

.

Answer:

\frac{1}{4}

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