$\frac{{{{\log }_2}a}}{3} = \frac{{{{\log }_2}b}}{4} = \frac{{{{\log }_2}c}}{{5{\rm{\lambda }}}}$ $and$ ${a^{ - 3}}{b^{ - 4}}c = 1$ then ${\rm{\lambda }} = $

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

\frac{{{{\log }_2}a}}{3} = \frac{{{{\log }_2}b}}{4} = \frac{{{{\log }_2}c}}{{5\mathrm{\lambda }}}

and

{a^{ - 3}}{b^{ - 4}}c = 1

then

\mathrm{\lambda } =

View answer and explanation

Correct option: C

Option C (5) is correct because it is the value of lambda that satisfies the given logarithmic equation and the condition

{a^{ - 3}}{b^{ - 4}}c = 1

Practice All Chapter MCQs →

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\frac{3}{4}