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Logarithms

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Question : 11 of 85

Marks: +1, -0

If

\log_{16} x + (\log_{16} x)^2 + (\log_{16} x)^3 + \cdots = \frac{1}{3}

x

is

2
4
$\frac{1}{2}$
$\frac{1}{4}$

Solution:

Concept:

Infinite geometric series

t + t^2 + t^3 + \cdots

converges to

\frac{t}{1-t}

if

|t| < 1

.

Explanation:

Let

t = \log_{16} x

.

The series becomes

t + t^2 + t^3 + \cdots = \frac{1}{3}

.

This is a geometric series with first term

t

and common ratio

t

.

For convergence,

|t| < 1

.

The sum is

\frac{t}{1-t}

.

Set

\frac{t}{1-t} = \frac{1}{3}

.

Cross-multiply:

3t = 1 - t \Rightarrow 4t = 1 \Rightarrow t = \frac{1}{4}

.

Thus

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

.

So

x = 16^{1/4} = (2^4)^{1/4} = 2

.

Check:

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

, so the series converges.

Answer:

x = 2

, option A.

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