Question

The value of \[6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right)\]is
A. 4
B. 5
C. 6
D. 9

Answer 1

Hint: We assume the value that is being repeated as a variable inside the square root and then use that variable to convert the remaining value in terms of that variable. Calculate the roots of the equation formed and select the positive root to substitute in the given equation. Use properties of logarithm to calculate the value of the equation.
* A quadratic equation is an equation with the highest power of the variable as 2. Roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
* \[\log {m^n} = n\log m\]

Complete step by step answer:
Let us assume \[x = \sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } \]
Then we can substitute this value of x inside the root as
\[ \Rightarrow x = \sqrt {4 - \dfrac{1}{{3\sqrt 2 }}x} \]
Square both sides of the equation
\[ \Rightarrow {x^2} = {\left( {\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}x} } \right)^2}\]
Cancel square root by square power in RHs of the equation
\[ \Rightarrow {x^2} = 4 - \dfrac{1}{{3\sqrt 2 }}x\]
Take LCM in RHS of the equation
\[ \Rightarrow {x^2} = \dfrac{{12\sqrt 2 - x}}{{3\sqrt 2 }}\]
Cross multiply the values from RHS to LHS
\[ \Rightarrow 3\sqrt 2 {x^2} = 12\sqrt 2 - x\]
Shift all values to LHS of the equation
\[ \Rightarrow 3\sqrt 2 {x^2} + x - 12\sqrt 2 = 0\] … (1)
Compare the equation with general quadratic equation i.e. \[a{x^2} + bx + c = 0\]
\[ \Rightarrow a = 3\sqrt 2 ;b = 1;c = - 12\sqrt 2 \]
Calculate the roots of equation (1) by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substitute the values of a, b and c in the formula
\[ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 3\sqrt 2 \times ( - 12\sqrt 2 )} }}{{2 \times 3\sqrt 2 }}\]
\[ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 12 \times 12 \times 2} }}{{6\sqrt 2 }}\]
\[ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 288} }}{{6\sqrt 2 }}\]
\[ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {289} }}{{6\sqrt 2 }}\]
\[ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{{17}^2}} }}{{6\sqrt 2 }}\]
Cancel square root by square power
\[ \Rightarrow x = \dfrac{{ - 1 \pm 17}}{{6\sqrt 2 }}\]
First value: \[x = \dfrac{{ - 1 - 17}}{{6\sqrt 2 }}\]
\[ \Rightarrow x = \dfrac{{ - 18}}{{6\sqrt 2 }}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow x = \dfrac{{ - 3}}{{\sqrt 2 }}\]
Second value: \[x = \dfrac{{ - 1 + 17}}{{6\sqrt 2 }}\]
\[ \Rightarrow x = \dfrac{{16}}{{6\sqrt 2 }}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow x = \dfrac{8}{{3\sqrt 2 }}\] … (2)
Now we use this value and substitute it in the given equation in the question
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}x} \right)\]
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }} \times \dfrac{8}{{3\sqrt 2 }}} \right)\]
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 + {\log _{3/2}}\left( {\dfrac{8}{{18}}} \right)\]
Cancel same factors from numerator and denominator in bracket
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 + {\log _{3/2}}\left( {\dfrac{4}{9}} \right)\]
Now we can take inverse of term in bracket
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 + {\log _{3/2}}{\left( {\dfrac{4}{9}} \right)^{ - 1}}\]
Use property of logarithm \[\log {m^n} = n\log m\]
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 - {\log _{3/2}}\left( {\dfrac{9}{4}} \right)\]
Write the terms in bracket in terms of square of a number
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 - {\log _{3/2}}{\left( {\dfrac{3}{2}} \right)^2}\]
Use property of logarithm \[\log {m^n} = n\log m\]
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 - 2{\log _{3/2}}\left( {\dfrac{3}{2}} \right)\]
Put the value of \[{\log _{3/2}}\left( {\dfrac{3}{2}} \right) = 1\] as the base is same as the number
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 6 - 2\]
\[ \Rightarrow 6 + {\log _{3/2}}\left( {\dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}\sqrt {4 - \dfrac{1}{{3\sqrt 2 }}.....} } } } \right) = 4\]

\[\therefore \]Correct option is A.

Note: Many students make the mistake of calculating the value of logarithm using a calculator when we obtain a term in fraction in RHS, keep in mind we try to cancel as many terms as possible to simplify log value using properties of logarithm.