Question

Solve the equations for $(x,y)$ :$(3x{)}^{\log 3}=(4y{)}^{\log 4},4^{\log x}=3^{\log y}$

• A. $\frac{1}{2},\frac{1}{3}$
• B. $\frac{1}{3},\frac{1}{4}$
• C. $\frac{1}{4},\frac{1}{3}$
• D. $\frac{1}{6},\frac{1}{8}$

Answer 1

Answer: (B)

$\frac{1}{3},\frac{1}{4}$

We have,
$(3x{)}^{\log 3}=(4y{)}^{\log 4}$ and $4^{\log x}=3^{\log y}$
take log on both sides,
$\log 3(\log 3+\log x)=\log 4(\log y+\log 4)\Rightarrow \left(1\right)$
$[\because \log a^m=m\log a\text{\&}\log (ab)=\log a+\log b]$
and
$\log x\log 4=\log 3\log y\Rightarrow \left(2\right)$
Now putting value of $\log y$ from (2) in equation (1) we get,
$\log 3(\log 3+\log x)=\log 4(\frac{\log 4}{\log 3}\log x+\log 4)$
$\Rightarrow {\log }^{2}3(\log 3+\log x)={\log }^{2}4(\log x+\log 3)$
$\Rightarrow ({\log }^{2}3-{\log }^{2}4)(\log x+\log 3)=0$
$\Rightarrow \log x+\log 3=0\Rightarrow \log x=-\log 3=\log \frac{1}{3}\Rightarrow x=\frac{1}{3}$
Now using (2), $\log y=\frac{\log 4}{\log 3}(-\log 3)=-\log 4=\log \frac{1}{4}\Rightarrow y=\frac{1}{4}$