Question

The value of the expression $6+{{\log }_{\dfrac{3}{2}}}\left( \dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}} \right)$ is

Answer 1

Hint: Assume $x=\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}}$. Multiply both sides by $3\sqrt{2}$ and square both sides and hence prove that $18{{x}^{2}}+x-4=0$. Solve the quadratic expression for x using the quadratic formula and hence find the roots of the quadratic equation. Remove extraneous roots and hence find the value of the expression $\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}}$. Hence find the value of $6+{{\log }_{\dfrac{3}{2}}}\left( \dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}} \right)$

Complete step by step answer:
Let $x=\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}}$
Multiplying both sides by $3\sqrt{2}$, we get
$3\sqrt{2}x=\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}}$
Squaring both sides, we get
$18{{x}^{2}}=4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}}$
Hence, we have
$18{{x}^{2}}=4-x$
Adding x on both sides, we get
$18{{x}^{2}}+x=4$
Subtracting 4 on both sides, we get
$18{{x}^{2}}+x-4=0$
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here a = 18, b = 1 and c=-4
Hence, we have
$x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\left( 18 \right)\left( -4 \right)}}{2\left( 18 \right)}$
Simplifying, we get
$x=\dfrac{-1\pm \sqrt{1+288}}{36}=\dfrac{-1\pm \sqrt{289}}{36}$
We know that $\sqrt{289}=17$
Hence, we have
$x=\dfrac{-1\pm 17}{36}$
Taking the positive sign, we get
$x=\dfrac{-1+17}{36}=\dfrac{4}{9}$
Taking the negative sign, we get
$x=\dfrac{-1-17}{36}=-\dfrac{1}{2}$
Since $\sqrt{x}\ge 0,\forall x\ge 0$, we have
$x=\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}}\ge 0$
Hence $x=\dfrac{-1}{2}$ is rejected.
Hence, we have $x=\dfrac{4}{9}$
Hence, we have
$6+{{\log }_{\dfrac{3}{2}}}\left( \dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}} \right)=6+{{\log }_{\dfrac{3}{2}}}\left( {{\left( \dfrac{3}{2} \right)}^{-2}} \right)$
We know that ${{\log }_{a}}{{a}^{n}}=n$

Hence, we have
$6+{{\log }_{\dfrac{3}{2}}}\left( \dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\cdots }}} \right)=6-2=4$

Note: In the equations involving logarithms, inverse trigonometric equations and square roots the most common mistake done by students is that they do not check the domain and range of these functions and hence end up taking an extraneous root in the solution which actually does not satisfy the equation. This can be avoided by substituting the value back and checking if that is the solution of the equation or not.