JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Medium

1 of 1

Application +4 | -1

The solution set of the equation

{\left[ {4\left( {1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{{27}} + ...} \right)} \right]^{{{\log }_2}x}} = {\left[ {54\left( {1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + ...} \right)} \right]^{{{\log }_x}2}}

View answer and explanation

Correct option: A

\left\{ {4,\;\frac{1}{4}} \right\}

Option A

\left\{{4,\;\frac{{1}}{{4}}}\right\}

is the correct solution set because it satisfies the given logarithmic equation after simplification using geometric series and logarithm properties.

More Questions from QUADRATIC EQUATIONS

Practice All Chapter MCQs →

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

\alpha ,\beta

{x^2} - 3x + a = 0,a \in R

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

\;a \ne 0,

9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)