Question

Find the value of


`\\ \rm\Rrightarrow {6+log_{(3)/(2)}\left((1)/(3√(2))\sqrt{4-(1)/(3√(2))\sqrt{4-(1)/(3√(2))\sqrt{4-(1)/(3√(2))\dots}}}\right)}`


Options are

`\sf \circ -4`

`\sf \circ 1`

`\sf \circ 4`

`\sf \circ 2`
Note:-

Kindly don't answer wrong if you don't know .

Spams/copied from web/short/wrong/irrelevant answers will be deleted on the spot .



​

Answer 1

`\sqrt{4 - \frac1{3\sqrt2} \sqrt{4 - \frac1{3\sqrt2} \sqrt{4 - \frac1{3\sqrt2} √(\cdots)}}}`

Starting from the identity

`(x - y)^2 = x^2 - 2xy + y^2`

take the positive square root on both sides.

`x - y = √(x^2 - 2xy + y^2)`

Note that we must have
`x\ge y`. Rewrite the radicand and substitute
`x-y`.

`x - y = √(x^2 - xy - y (x - y)) \\\\ ~~~~ = \sqrt{x^2 - xy - y √(x^2 - xy - y (x - y))} \\\\ ~~~~ = \sqrt{x^2 - xy - y \sqrt{x^2 - xy - y √(x^2 - xy - y (x - y))}} \\\\ ~~~~ \vdots \\\\ ~~~~ = \sqrt{x^2 - xy - y \sqrt{x^2 - xy - y \sqrt{x^2 - xy - y √(\cdots)}}}`

Let
`y=\frac1{3\sqrt2}`. Solve for
`x`.

`x^2 - \frac x{3\sqrt2} = 4 \\\\ x^2 - \frac x{3\sqrt2} + \frac1{72} = (289)/(72) \\\\ \left(x - \frac1{6\sqrt2}\right)^2 = (289)/(72) \\\\ x - \frac1{6\sqrt2} = \pm (17)/(6\sqrt2) \\\\ x = (18)/(6\sqrt2) \text{ or } x = -(16)/(6\sqrt2) \\\\ x = \frac3{\sqrt2} \text{ or } x = -\frac8{3\sqrt2}`

Take the positive solution to ensure
`x>y`. Then the infinitely nested root expression in the logarithm converges to

`x - y = \frac3{\sqrt2} - \frac1{3\sqrt2} = \frac{4\sqrt2}3`

and the overall expression has a value of

`6 + \log_(\frac32) \left(\frac1{3\sqrt2} * \frac{4\sqrt2}3\right) = 6 + \log_(\frac32) \left(\frac49\right) \\\\ ~~~~ = 6 + \log_(\frac32) \left(\frac23\right)^2 \\\\ ~~~~ = 6 - 2 \log_(\frac32) \left(\frac32\right) \\\\ ~~~~ = 6 - 2 = \boxed{4}`

Answer 2

4

Explanation:

Given,

`6+log(3)/(2) ((1)/(3√(2) ) \sqrt{4-(1)/(3√(2) )\sqrt{4-(1)/(3√(2) ) ...} }`

Let,

`x= \sqrt{4-(1)/(3√(2) )\sqrt{4-(1)/(3√(2) )\\`

By this we get

`x=\sqrt{4-(1)/(3√(2) )(x) } }`

On squaring both sides,

`x^(2) =4-(1)/(3√(2) )(x) } }\\\\x^(2) -4+(x)/(3√(2) ) =0\\\\3√(2) x^(2) -12√(2) +x=0\\\\x=\frac{-1+\sqrt{1-(-12√(2))*(3√(2))*4 } }{2*3√(2) } \\\\x=(-1+√(289) )/(6√(2) ) \\\\x=(-1+17)/(6√(2) ) \\\\x=(8)/(3√(2) )`

Now,

`6+log(3)/(2) [(1)/(3√(2) ) *(8)/(3√(2) ) ]+log(3)/(2) *[(8)/(9*2) ]\\\\6+log(3)/(2)((4)/(9) )\\\\6-log(3)/(2)((9)/(4) )\\\\6-log(3)/(2) ((3)/(2))^2 \\\\6-2=4`