log-10-x-log-10-x-8log-10-y-log-10-2-x-log-10-2-y-3-log-10-y-8log-10-x-log-10-y-log-10-2-x-log-10-2-y-0-find-x-amp-y-

Question Number 84459 by jagoll last updated on 13/Mar/20

{\begin{cases} \log_{10} (x) + \frac{\log_{10} (x) + {8 \log}_{10} (y)}{\log_{10}^{2} (x) + \log_{10}^{2} (y)} = 3 \\ \log_{10} (y) + \frac{{8 \log}_{10} (x) - \log_{10} (y)}{\log_{10}^{2} (x) + \log_{10}^{2} (y)} = 0 \end{cases}

find x & y

Commented by john santu last updated on 13/Mar/20

let \log_{10} (x) = a & \log_{10} (y) = b

(1) a + \frac{a + 8 b}{a^{2} + b^{2}} = 2 \land (2) b + \frac{8 a - b}{a^{2} + b^{2}} = 0

b (a + \frac{a + 8 b}{a^{2} + b^{2}}) = 2 b (1)

a (b + \frac{8 a - b}{a^{2} + b^{2}}) = 0 (2)

\Rightarrow 2 ab + \frac{{8 a}^{2} + {8 b}^{2}}{a^{2} + b^{2}} = 2 b

Commented by jagoll last updated on 13/Mar/20

2 ab + 8 = 2 b \Rightarrow 4 = b - ab

b = \frac{4}{1 - a} \land a^{2} b + b^{3} + 8 a - b = 0

thank you sir . i have idea to

solving this problem

Answered by MJS last updated on 13/Mar/20

\log_{10} x = u \land \log_{10} y = v

v = p u \land p, u \neq 0

{\begin{cases} u - 3 + \frac{8 p + 1}{(p^{2} + 1) u} = 0 \\ p u - \frac{p - 8}{(p^{2} + 1) u} = 0 \end{cases}

{\begin{cases} u^{2} - 3 u + \frac{8 p + 1}{p^{2} + 1} = 0 \\ u^{2} - \frac{p - 8}{p (p^{2} + 1)} = 0 \end{cases}

(2) - (1) \Rightarrow u = \frac{2 (4 p^{2} + p - 4)}{3 p (p^{2} + 1)}

insert in (1) or (2) \Rightarrow

p^{4} + \frac{104}{55} p^{3} - \frac{133}{55} p^{2} + \frac{8}{11} p + \frac{64}{55} = 0

(p^{2} + \frac{4 - (13 + \sqrt{1049})}{55} p + \frac{123 + \sqrt{1049}}{110}) (p^{2} + \frac{4 (13 - \sqrt{1049})}{55} p + \frac{123 - \sqrt{1049}}{110}) = 0

we can exactly solve this

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