The value of 6 + log(3/2)

Question

The value of 

6 + \(\text{log}_{\frac{3}{2}}\)\(\bigg[\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}}}}}...\) is

(a) \(\frac{8}{3\sqrt2}\)

(b) \(\frac{4}{3}\)

(c) 8 

(d) 4

Answer 1

(d) 4

Let A  = 6 + \(\text{log}_{\frac{3}{2}}\)\(\bigg[\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}}}}}...\) 

Let P = \(\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}\sqrt{4-\frac{1}{3\sqrt2}}}}}...\)

P = \(\sqrt{4-\frac{1}{3\sqrt2}P}\) ⇒ P² = 4 - \(\frac{1}{3\sqrt2}\)P ⇒ P² + \(\frac{1}{3\sqrt2}\)P - 4 = 0

⇒ P = \(\frac{-\frac{1}{3\sqrt2}±\sqrt{\frac{1}{18}+16}}{2}\) = \(\frac{-\frac{1}{3\sqrt2}±\frac{17}{3\sqrt2}}{2}\)

(Applying the formula for roots of Q.E.)

⇒ P = \(\frac{16}{3\times2\sqrt2}\) or \(\frac{-18}{3\times2\sqrt2}\) = \(\frac{8}{3\sqrt2}\) or \(-\frac{3}{\sqrt2}\)

Neglecting p = \(\frac{-3}{\sqrt2}\) as p > 0, we have p = \(\frac{8}{3\sqrt2}\)

∴ A = 6 + \(\text{log}_{\frac{3}{2}}\)\(\bigg(\frac{1}{3\sqrt2}\times\frac{8}{3\sqrt2}\bigg)\) = 6 + \(\text{log}_{\frac{3}{2}}\) \(\big(\frac{4}{9}\big)\)

= 6 + \(\text{log}_{\frac{3}{2}}\) \(\big(\frac{3}{2}\big)^{-2}\) = 6 - 2 \(\text{log}_{\frac{3}{2}}\) \(\frac{3}{2}\) = 6 - 2 = 4.