JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

There exist a positive number

\lambda

, such that

\log _2 x+\log _4 x+\log _8 x=\log _\lambda x

, for all positive real numbers

x

.

If

\lambda=\sqrt[b]{a}

, where

a, b \in N

, the smallest possible value of

(a+b)

is equal to

View answer and explanation

Correct option: D

Option D (75) is correct because simplifying the logarithmic equation leads to

\lambda = 2^{6/11}

, and expressing this in the form

\lambda = \sqrt[b]{a}

gives

a = 2^6 = 64

and

b = 11

, so

a + b = 75

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