Question

Solve the following equations for $x$ and $y$. ${\log }_{10}x+\frac{1}{2}{\log }_{10}x+\frac{1}{4}{\log }_{10}x+...=y$ and $\frac{1+3+5+....+(2y-1)}{4+7+10+....+(3y+1)}=\frac{20}{7{\log }_{10}x}$

• A. $x={10}^4$ and $y=8$
• B. $x={10}^5$ and $y=10$
• C. $x={10}^3$ and $y=6$
• D. $x={10}^6$ and $y=12$

Answer 1

Answer: (B)

$x={10}^5$ and $y=10$

From the first equation,
${\log }_{10}x\{1+\frac{1}{2}+\frac{1}{4}+..\infty \}=y$
$\Rightarrow {\log }_{10}x\left\{\frac{1}{1-1/2}\right\}=y$
$\Rightarrow 2{\log }_{10}x=y\cdots \left(1\right)$
From the second equation,
$\Rightarrow \frac{\frac{y}{2}\{1+2y-1\}}{\frac{y}{2}\{4+3y+1\}}=\frac{20}{7{\log }_{10}x}$
$\Rightarrow \frac{2y}{3y+5}=\frac{20}{7{\log }_{10}x}$
$\Rightarrow 7y\left(2{\log }_{10}x\right)=60y+100$
$\Rightarrow 7y\left(y\right)=60y+100$ from $\left(1\right)$
$\Rightarrow 7y^2-60y-100=0$
$\therefore (y-10)(7y+10)=0$
$\Rightarrow y=10,y\ne \frac{-10}{7}(\because yϵI_{+})$
from $\left(1\right)$, $2{\log }_{10}x=10$
$\Rightarrow {\log }_{10}x=5$
$\therefore x={10}^5$
Hence required solution is,
$x={10}^{5}andy=10$