Question

What will be the value of $6 + {\log _{\dfrac{3}{2}}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right)$?
$(a){\text{ 4}} \\\ (b){\text{ 5}} \\\ (c){\text{ 6}} \\\ (d){\text{ 9}} \\\$

Answer 1

Hint – In this question assume that $y = \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } }$, and then again we can see that the expression can be simplified $y = \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - y}$. Solve this to find the value of y and use them along with basic logarithmic identities to get the answer.

Complete step-by-step solution -
Given equation is
$6 + {\log _{\dfrac{3}{2}}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right)$............................. (1)
Let $y = \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } }$
As this is infinite going series so this series is also written as
$\Rightarrow y = \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - y}$
$\Rightarrow 3\sqrt 2 y = \sqrt {4 - y}$
Now squaring on both sides we have,
$\Rightarrow 18{y^2} = 4 - y$
$\Rightarrow 18{y^2} + y - 4 = 0$
Now this is a quadratic equation so apply quadratic formula which is
$y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ Where (a = 18, b = 1 and c = -4).
$\Rightarrow y = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( {18} \right)\left( { - 4} \right)} }}{{2\left( {18} \right)}} = \dfrac{{ - 1 \pm \sqrt {1 + 288} }}{{36}} = \dfrac{{ - 1 \pm \sqrt {289} }}{{36}} = \dfrac{{ - 1 \pm 17}}{{36}}$
$y = \dfrac{{ - 18}}{{36}},\dfrac{{16}}{{36}}$
$\Rightarrow y = \dfrac{{ - 1}}{2},\dfrac{4}{9}$
So from equation (1) we have,
$\Rightarrow 6 + {\log _{\dfrac{3}{2}}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 + {\log _{\dfrac{3}{2}}}\left( y \right)$
So as we see y cannot be negative because negative log is undefined.
Therefore possible case is y = (4/9)
$\Rightarrow 6 + {\log _{\dfrac{3}{2}}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 + {\log _{\dfrac{3}{2}}}\left( {\dfrac{4}{9}} \right)$
Now as we know ${\log _a}b = \dfrac{{\log b}}{{\log a}},\log {k^2} = 2\log k{\text{ and }}\log \dfrac{c}{d} = - \log \dfrac{d}{c}$ so use this property in above equation we have,
$\Rightarrow 6 + {\log _{\dfrac{3}{2}}}\left( {\dfrac{4}{9}} \right) = 6 + \dfrac{{\log {{\left( {\dfrac{2}{3}} \right)}^2}}}{{\log \dfrac{3}{2}}} = 6 + \dfrac{{2\log \dfrac{2}{3}}}{{ - \log \dfrac{2}{3}}} = 6 - 2 = 4$
So this is the required answer.
Hence option (A) is correct.

Note – Such type of questions are always pattern based, there is indulgence of one pattern than keeps on repeating inside itself. So the basic approach to all such problems is simply to assume the basic repeating pattern as a variable. Basic logarithmic identities like ${\log _a}b = \dfrac{{\log b}}{{\log a}},\log {k^2} = 2\log k{\text{ and }}\log \dfrac{c}{d} = - \log \dfrac{d}{c}$ are advised to be remembered as it helps solving problems of this kind.