Question

anyone can solve this problem​

Answer 1

Required Answer:-

Given to find:

$\sf \mapsto 6 + \log_{^{3}/_{2} } \bigg(\dfrac{1}{3 \sqrt{2} } \sqrt{4 - \dfrac{1}{3 \sqrt{2} } \sqrt{4 - \dfrac{1}{3 \sqrt{2} }.... } } \bigg)$

Solution:

Let us assume that,

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

Squaring both sides, we get,

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

As mentioned earlier that,

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

So,

$\sf \implies {x}^{2} = 4 - \dfrac{1}{3 \sqrt{2} }x$

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

Multiplying both sides bt 3√2, we get,

$\sf \implies 3 \sqrt{2} {x}^{2} + x - 12 \sqrt{2} = 0$

Now, we will solve the quadratic equation.

Here,

$\sf a = 3 \sqrt{2}$

$\sf b = 1$

$\sf c = - 12 \sqrt{2}$

By quadratic formula,

$\sf x_{1,2} = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Putting all the values here, we get,

$\sf x = \dfrac{ - 1 \pm \sqrt{ {1}^{2} - 4 \times 3 \sqrt{2} \times ( - 12 \sqrt{2} ) } }{2 \times 3 \sqrt{2} }$

$\sf x = \dfrac{ - 1 \pm \sqrt{1+ 288} }{2 \times 3 \sqrt{2} }$

$\sf x = \dfrac{ - 1 \pm 17}{6\sqrt{2} }$

So,

$\sf x_{1} = \dfrac{ - 1 + 17}{6 \sqrt{2} } = \dfrac{16}{6 \sqrt{2} } = \dfrac{8}{3 \sqrt{2} }$

Again,

$\sf x_{2} = \dfrac{ - 1 - 17}{6 \sqrt{2} } = \dfrac{ - 18}{6 \sqrt{2} } = \dfrac{ - 3}{ \sqrt{2} }$

But,

$\sf \mapsto \dfrac{ - 3}{ \sqrt{2} } < 0$

So, we omit this value.

So,

$\sf x = \dfrac{8}{3 \sqrt{2} }$

Now,

$\sf6 + \log_{^{3}/_{2} } \bigg(\dfrac{1}{3 \sqrt{2} } \sqrt{4 - \dfrac{1}{3 \sqrt{2} } \sqrt{4 - \dfrac{1}{3 \sqrt{2} }.... } } \bigg)$

$\sf = 6 + \log_{^{3}/_{2} } \bigg(\dfrac{1}{3 \sqrt{2} } \times \dfrac{8}{3 \sqrt{2} } \bigg)$

$\sf = 6 + \log_{^{3}/_{2} } \bigg(\ \dfrac{8}{9 \times 2} \bigg)$

$\sf = 6 + \log_{^{3}/_{2} } \bigg(\ \dfrac{4}{9 } \bigg)$

$\sf = 6 + \log_{^{3}/_{2} } \bigg(\ \dfrac{2}{3} \bigg)^{2}$

$\sf = 6 + \log_{^{3}/_{2} } \bigg(\ \dfrac{3}{2} \bigg)^{ - 2}$

$\sf = 6 + - 2 \cancel{\log_{^{3}/_{2} } \bigg(\ \dfrac{3}{2} \bigg)}$

$\sf = 6 - 2$

$\sf = 4$

Hence, the required answer is 4.

Answer 2

Answer:

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