Question

$\textsf{Find the value of: }\\\\\displaystyle 6 + log_{\frac{3}{2}}\Bigg[\dfrac{1}{3\sqrt{2}}\sqrt{4 - \dfrac{1}{3\sqrt2}\sqrt{4 - \dfrac{1}{3\sqrt2}\sqrt{4 - \dfrac{1}{3\sqrt2}\sqrt{4 - \dfrac{1}{3\sqrt2}}}}}~~~~\Huge\Bigg]$

Answer 1

Let $\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}.....}}$ be x .

$x=\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}}.....$

Then squaring both sides , we get :

$x^2=4-\dfrac{1}{3\sqrt{2}}(\sqrt{4-\dfrac{1}{3\sqrt{2}}\sqrt{4-\dfrac{1}{3\sqrt{2}}}.....)$

$\implies x^2=4-\dfrac{1}{3\sqrt{2}}(x)\\\\\implies 3\sqrt{2}x^2=12\sqrt{2}-x\\\\\implies 3\sqrt{2}x^2+x-12\sqrt{2}=0$

By Sri dharacharya formula :

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\implies x=\dfrac{-1\pm\sqrt{1+288}}{6\sqrt{2}}\implies x=\dfrac{-1\pm17}{6\sqrt{2}}$

If x becomes negative , then the entire thing will become a complex number . So neglect the value of x when negative .

$x=\dfrac{-1+17}{6\sqrt{2}}\implies x=\dfrac{16}{6\sqrt{2}}\\\\\implies x=\dfrac{8}{3\sqrt{2}}\\\\\implies x=\dfrac{8\sqrt{2}}{3\times 2}\\\\\implies x=\dfrac{4\sqrt{2}}{3}$

Now just put the value of x in the equation .

$6+log_{\dfrac{3}{2}}(\dfrac{1}{3\sqrt{2}}\times \dfrac{4\sqrt{2}}{3})\\\\\implies 6+\dfrac{4}{9}(log_{\dfrac{3}{2}})\\\\\implies 6+(\dfrac{3}{2})^{-2}(log_{\dfrac{3}{2}})\\\\\implies 6+(-2)(log_{\dfrac{3}{2}})^{\dfrac{3}{2}}\\\\\implies 6-2 \implies 4$

The answer is 4 .

Sorry for lengthy method .

Formulas used :

$log_aa=1\\\\log a^b=bloga\\\\(\dfrac{a}{b})^{-m}=\dfrac{b^m}{a^m}$

Answer 2

Answer :-----

formula :

$\begin{lgathered}x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\implies x=\dfrac{-1\pm\sqrt{1+288}}{6\sqrt{2}}\implies x=\dfrac{-1\pm17}{6\sqrt{2}}\end{lgathered}$

value of x

$\begin{lgathered}6+log_{\dfrac{3}{2}}(\dfrac{1}{3\sqrt{2}}\times \dfrac{4\sqrt{2}}{3})\\\\\implies 6+\dfrac{4}{9}(log_{\dfrac{3}{2}})\\\\\implies 6+(\dfrac{3}{2})^{-2}(log_{\dfrac{3}{2}})\\\\\implies 6+(-2)(log_{\dfrac{3}{2}})^{\dfrac{3}{2}}\\\\\implies 6-2 \implies 4\end{lgathered} </p><p>6+log237$

( 3 21 × 34 2)

⟹6+ 94(log 23)

⟹6+( 23 ) −2 (log 23 )

⟹6+(−2)(log 23 ) 23

⟹//6−2//⟹4