Question

If ${\log }_2(x+y)={\log }_3(x-y)=\frac{\log 25}{\log 0.2}$. Find the values of $x$ and $y$.

Answer 1

Que :If ${\log }_2(x+y)={\log }_3(x-y)=\frac{\log 25}{\log 0.2}$.

$\to lo{g}_2(x+y)=\frac{lo{g}_e(x+y)}{lo{g}_{e}2}$($\because$ using $lo{g}_{e}b=\frac{lo{g}_{e}b}{lo{g}_{e}a})$

$lo{g}_3(x-y)=\frac{lo{g}_e(x-y)}{lo{g}_{e}3}$

$\frac{log25}{log0.2}=\frac{log25}{log\frac{2}{10}}=\frac{log25}{log2/5}=\frac{log{5}^2}{log{5}^2}$

$=\frac{2log5}{-2log5}=-2$

$\to$ so,

$\therefore \therefore \frac{lo{g}_e(x+y)}{lo{g}_{e}2}=-2$

$\therefore lo{g}_e(x+y)=-2lo{g}_{e}2$

$\therefore lo{g}_e(x+y)=lo{g}_e{2}^{-2}$

$\therefore x+y=2^{-2}$

$\therefore x+y=\frac{1}{4}...\left(i\right)$

$\therefore \frac{lo{g}_e(x-y)}{lo{g}_{e}3}=-2$

$\therefore lo{g}_e(x-y)=-2lo{g}_{e}3$

$\therefore lo{g}_e(x-y)=lo{g}_e{3}^2$

$\therefore x-y=3^{-2}=2/3^2$

$\therefore x-y=\frac{2}{3}...\left(ii\right)$

$\to$ adding equation (i) & (ii)

$x+y=2/4$

$+x-y=2/9$

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$2x=\frac{1}{4}+\frac{1}{9}$

$\therefore 2x=\frac{9+4}{36}$

$\therefore 2x=\frac{13}{36}$

$\therefore x=\frac{13}{72}$

value of x in equation

(i) we get

$\frac{13}{72}+y=\frac{1}{4}$

$\therefore y=\frac{1}{4}-\frac{13}{72}$

$\therefore y=\frac{18-13}{72}$

$\therefore y=\frac{5}{72}$