Question

If equations ${\log }_{128}x+{\log }_{125}y=5$ and ${\log }_{y}125-{\log }_{x}128=1$ is satisfied by ordered pairs $(x_1$, $y_1)$ and $\left(x_2,y_2\right)$, then find the value of ${\log }_{21}\left({\log }_{10}{x}_1{x}_2{y}_1{y}_2\right)$.

Answer 1

Answer: 1

Let ${\log }_{128}x$ and ${\log }_{125}y=b$
$\therefore a+b=5\Rightarrow b=5-a$
$\text{ and }\frac{1}{b}-\frac{1}{a}=1\Rightarrow \frac{a-b}{ab}=1\Rightarrow a-b=ab$
$\Rightarrow a-(5-a)=a(5-a)\Rightarrow a^2-3a-5=0$
$\therefore a_1+a_2=3$
$\Rightarrow {\log }_{128}{x}_1+{\log }_{128}{x}_2=3\Rightarrow x_1{x}_2=(128{)}^3=2^{21}$
$\Theta b=5-a$
$\therefore b_1+b_2=10-\left(a_1+a_2\right)=7$
$\Rightarrow {\log }_{125}\left(y_1{y}_2\right)=7\Rightarrow y_1{y}_2=5^{21}$
$\therefore x_1{x}_2{y}_1{y}_2=10^{21}$
$\Rightarrow {\log }_{10}\left(x_2{y}_1{y}_2\right)=21$
$\Rightarrow {\log }_{21}\left({\log }_{10}{x}_1{x}_2{y}_1{y}_2\right)=1$