Question

$\sf{If \: x,y > 0, log_{x}(y) \: + \: log_{y}(x) \: = \dfrac{10}{3} \: and \: xy \: = \: 144, \: then \: \frac{x \: + \: y}{2} \: = }$

Please Answer This Question !!!


Spammers Be Away !!!


Chapter : Logarithm



​

Answer 1

$\mathfrak{\underline{\underline{\mathfrak{ \large{Solution : }}}}}$

Given,

$\mathsf{ \implies log_x \: y \: + \: log_y \: x \: = \: \dfrac{10}{3}} \\ \\ \mathsf{ \implies \: log_x \: y \: + \: \dfrac{1}{log_x \: y } \: = \: \dfrac{10}{3}} \\ \\ </p><p> \mathsf{ \implies \: \dfrac{( log_x \: y)^{2} \: + \: 1}{log_x \: y } \: = \: \dfrac{10}{3}}$

$\sf{Let,} \: \sf log_x \: y = \: z$

$\sf \implies \: \dfrac{z^{2} \: + \: 1}{z} \: = \: \dfrac{10}{3} \\ \\ \sf \implies \: 3 ({z}^{2} \: + \: 1) \: = \: 10z \\ \\ \sf \implies \: 3 {z}^{2} \: + \: 3 \: = \: 10z \\ \\ \sf \implies3 {z}^{2} \: - \: 10z \: + \: 3 \: = \: 0 \\ \\ \sf \implies3 {z}^{2} \: - \: 9z \: - \: z \: + \: 3 \: = \: 0 \\ \\ \sf \implies3z(z \: - 3) \: - \: 1(z \: - \: 3) \: = \: 0 \\ \\ \sf \implies(z \: - \: 3)(3z \: - \: 1) \: = \: 0$

By Zero Product Rule :

$\sf \therefore \: \: z \: = \: 3 \: and \: \dfrac{1}{3}$

Case 1:

$\sf \implies \: z \: = \: 3 \\ \\ \sf \implies \: log_x \: y \: = \: 3 \\ \\ \sf \implies {x}^{3} \: = \: y \: \: \: \: .......(1)$

According to question :

$\sf\implies xy \: = \: 144 \\ \\ \sf \therefore \: \: y\: = \: \dfrac{144}{x} \: \: \: \: \: .......(2)$

Put the value of y in (1),

$\sf \implies {x}^{3} \: = \: y \\ \\ \sf \implies {x}^{3} \: = \: \dfrac{144}{x} \\ \\ \sf \implies {x}^{4} \: = \: 144 \\ \\ \sf \implies x \: = \sqrt{144} \\ \\ \sf \therefore \: \: x \: = \: \sqrt{12}$

Put the value of x in (2),

$\sf\implies y \: = \: \dfrac{144}{x} \\ \\ \sf \implies y \: = \: \dfrac{144}{ \sqrt{12} } \\ \\ \sf \therefore \: \: y \: = \: 12 \sqrt{12}$

To Find :

$\sf = \: \dfrac{x \: + \: y}{2} \\ \\ \sf = \: \dfrac{ \sqrt{12} \: + \: 12 \sqrt{12} }{2} \\ \\ \sf = \: \dfrac{13 \sqrt{12} }{2} \\ \\ \sf = \: \dfrac{13 \: \times \: 2 \sqrt{3} }{2} \\ \\ \sf \: = \: 13 \sqrt{3}$

Case 2:

$\sf \implies \: z \: = \: 3 \\ \\ \sf \implies \: log_y\: x \: = \: 3 \\ \\ \sf \implies {y}^{3} \: = \: x \: \: \: \: .......(1)$

According to question :

$\sf\implies xy \: = \: 144 \\ \\ \sf \therefore \: \: y\: = \: \dfrac{144}{x} \: \: \: \: \: .......(2)$

Put the value of x in (1),

$\sf \implies {y}^{3} \: = \: x\\ \\ \sf \implies {y}^{3} \: = \: \dfrac{144}{y} \\ \\ \sf \implies {y}^{4} \: = \: 144 \\ \\ \sf \implies y\: = \sqrt{144} \\ \\ \sf \therefore \: \: y \: = \: \sqrt{12}$

Put the value of y in (2),

$\sf\implies y \: = \: \dfrac{144}{x} \\ \\ \sf \implies \sqrt{12} \: = \: \dfrac{144}{ x} \\ \\ \sf \implies x \: = \: \dfrac{144}{ \sqrt{12} } \\ \\ \sf \therefore \: \: x \: = \: 12 \sqrt{12}$

To Find :

$\sf = \: \dfrac{x \: + \: y}{2} \\ \\ \sf = \: \dfrac{ 12 \sqrt{12} \: + \: \sqrt{12} }{2} \\ \\ \sf = \: \dfrac{13 \sqrt{12} }{2} \\ \\ \sf = \: \dfrac{13 \: \times \: 2 \sqrt{3} }{2} \\ \\ \sf \: = \: 13 \sqrt{3}$

Answer 2

hope it helps......................