Question

[ Logarithm ]
Solve the given question ​

Answer 1

$\boxed{\textsf{Answer}}$

$\text{\underline{Step A: Finding the Power}}$

First, let's evaluate the power.

Let,

$\hookrightarrow A=\dfrac{1}{3\sqrt{2}},\ B=4$

Then we need to find

$\hookrightarrow 6+\log_{\frac{3}{2}}(A\sqrt{B-A\sqrt{B-A\sqrt{B\cdots} } } )$

So let

$\hookrightarrow X=A\sqrt{B-A\sqrt{B-A\sqrt{B\cdots } } }$.

By squaring both sides,

$\hookrightarrow X^{2}=A^{2}(B-A\sqrt{B-A\sqrt{B-A\sqrt{B\cdots } } } )$

$\hookrightarrow X^{2}=A^{2}(B-X)$

$\hookrightarrow X^{2}+A^{2}X-A^{2}B=0$

Let's solve the quadratic equation. Substitute the values of $A$ and $B$ we get,

$\hookrightarrow X^{2}+\dfrac{1}{18} X-\dfrac{2}{9} =0$

$\hookrightarrow 18X^{2}+X-4=0$

$\hookrightarrow (9X-4)(2X+1)=0$

$\hookrightarrow X=\dfrac{4}{9} \ \text{or}\ X=-\dfrac{1}{2}$

The second solution gets rejected, as square roots are non-negative.

Hence, the value of the power is $X=\dfrac{4}{9}$.

$\text{\underline{Step B: Using Logarithm Rules}}$

The logarithm,

$\hookrightarrow 6+\log_{\frac{3}{2} }\dfrac{4}{9}$

$\hookrightarrow 6+\log_{\frac{3}{2}}\left(\dfrac{3}{2} \right)^{-2}$

$\hookrightarrow 6-2\ \text{[Exponent Rule of Logarithm]}$

$\hookrightarrow 4 \ \text{\underline{(ANSWER.)}}$

$\boxed{\textsf{Conclusion}}$

Hence, the required answer is $4$.