Question

Given equations are $2^x\cdot 5^y=1$ and $5^{x+1}.2^y=2$, then the value of $x$ and $y$ are

• A. $x={\log }_{10}5$ and $y={\log }_{10}2$
• B. $x={\log }_{10}2$and $y={\log }_{10}5$
• C. $x={\log }_{10}\left(\begin{array}{c}\frac{1}{5}\end{array}\right)$and$y={\log }_{10}2$
• D. $x={\log }_{10}5$and$y={\log }_{10}\left(\begin{array}{c}\frac{1}{2}\end{array}\right)$

Answer 1

Answer: (C)

$x={\log }_{10}\left(\begin{array}{c}\frac{1}{5}\end{array}\right)$and$y={\log }_{10}2$

Given equations $2^x.5^y=1$
Taking logarithm on both sides,
${\log }_{10}\left(2^x{5}^y\right)={\log }_{10}1$
$\Rightarrow x{\log }_{10}2+y{\log }_{10}5=0$ .....(1)
Another equation $5^{x+1}.2^y=2$
$5^{x+1}=2^{1-y}$
Taking logarithm on both sides,
${\log }_{10}\left(5^{x+1}\right)={\log }_{10}\left(2^{1-y}\right)$
$\Rightarrow (x+1){\log }_{10}5=(1-y){\log }_{10}2$
$\Rightarrow x{\log }_{10}5+y{\log }_{10}2={\log }_{10}2-{\log }_{10}5$
$\Rightarrow x{\log }_{10}5+y{\log }_{10}2={\log }_{10}\left(\frac{2}{5}\right)$ ....(2)
Multiplying (1) by ${\log }_{10}2$ and (2) by ${\log }_{10}5$,and subtracting, we get
$x\left[\right({\log }_{10}{)}^2-\left({\log }_{10}{)}^2\right]=-{\log }_{10}5{\log }_{10}\left(\frac{2}{5}\right)$
$\Rightarrow x\left[{\log }_{10}\right(\frac{2}{5}\left){\log }_{10}10\right]=-{\log }_{10}5{\log }_{10}\left(\frac{2}{5}\right)$
$\Rightarrow x=-{\log }_{10}5$
$\Rightarrow x={\log }_{10}\left(\frac{1}{5}\right)$
Put this value in (1), we get
$y={\log }_{10}2$