Question

If $(x_1,y_1)$ and $(x_2,y_2)$ are the solution of the system of equation
${\log }_{225}\left(x\right)+{\log }_{64}\left(y\right)=4$
${\log }_{225}x-{\log }_y\left(64\right)=1$,
then find the value of ${\log }_{30}\left(x_1{y}_1{x}_2{y}_2\right).$

Answer 1

$let{\log }_{225}x=Pand{\log }_{64}y=Q$

$P+Q=4=>P=4-Q$

$andP-\frac{1}{Q}=1$

$PQ=1+Q$

$(4-Q)Q=1+Q$

$4Q-Q^2=1+Q$

$Q^2-3Q+1=0$

$productofroots=1,sum=3$ Now, $P+Q=4$

$Q=4-PPQ=1+Q$

$P(4-P)=1+4-P$

$4P-P^2=5-P$

$P^2-5P+5=0$

$productofroots=5,sum=5$

$Thus,{\log }_{64}{y}_1+{\log }_{64}{y}_2=3$

$y_1{y}_2={64}^3$

$and{\log }_{225}{x}_1+{\log }_{225}{x}_2=5$

$y_1{y}_2={64}^3$

$and{\log }_{225}{x}_1+{\log }_{225}{x}_2=5$

$x_1{x}_2={\left(225\right)}^5$

$Now,$

$x_1{x}_2{y}_1{y}_2={64}^3.{\left(225\right)}^5={\left(2\right)}^3.{\left(15\right)}^{10}$

$x_1{x}_2{y}_1{y}_2=2^{16}.3^{10}.5^{10}=2^8.{30}^{10}$

${\log }_{30}\left(x_1{x}_2{y}_1{y}_2\right)=8{\log }_{30}2+10$