Given-2log-5-x-2-log-5-y-1-2-0-log-5-x-2-2x-2-0-find-the-value-of-y-

Question Number 105994 by bemath last updated on 02/Aug/20

G i v e n {\begin{cases} 2 \log_{5} (x + 2) - \log_{5} (y) + \frac{1}{2} = 0 \\ \log_{5} (x^{2} + 2 x - 2) = 0 \end{cases}

f i n d t h e v a l u e o f y

Commented by PRITHWISH SEN 2 last updated on 02/Aug/20

x^{2} + 2 x - 2 = 1 \Rightarrow x = 1, - 3

\frac{{(x + 2)}^{2}}{y} = \frac{1}{\sqrt{5}} \Rightarrow x = 1, y = 9 \sqrt{5}, x = - 3, y = \sqrt{5}

(x, y) = (1, 9 \sqrt{5}), (- 3, \sqrt{5})

Answered by bobhans last updated on 02/Aug/20

(1) x^{2} + 2 x - 2 = 1 \Rightarrow x^{2} + 2 x - 3 = 0 \land x > - 2

\Rightarrow (x + 3) (x - 1) = 0 \Rightarrow x = 1, x = - 3 (rejected)

(2) 2 \log_{5} (x + 2) - \log_{5} (y) = - \frac{1}{2}

\log_{5} (\frac{{(x + 2)}^{2}}{y}) = \log_{5} (\frac{1}{\sqrt{5}})

\to y = \sqrt{5} {(x + 2)}^{2} = 9 \sqrt{5} =

Facebook Tweet Pin