Question

If $(3x{)}^{\log 3}=(4y{)}^{\log 4}$ and $4^{\log x}=3^{\log y}$, then the value of $x$ will be

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $3$
• D. $4$

Answer 1

The correct option is B $\frac{1}{3}$

Taking $\log$ on both sides of given equations,
$\log 3\log \left(3x\right)=\log 4\log \left(4y\right)$
$\log 3(\log 3+\log x)=\log 4(\log 4+\log y)[\because \log (ab)=\log a+\log b]$ $\Rightarrow$ (i)
And, $\log x\log 4=\log y\log 3$
$\log x\left(\log 4\right)=\left(\log y\right)\log 3$ $\Rightarrow$ (ii)
From (i)
$(\log 4{)}^2-(\log 3{)}^2=\log 3\log x-\log 4\log y$ $\Rightarrow$ (iii)
From (ii)
$\frac{\log x}{\log 3}=\frac{\log y}{\log 4}=\lambda \left(say\right)$ $\Rightarrow$ (iv)
From (iv) and (iii)
$(\log 4{)}^2-(\log 3{)}^2=\left[(\log 3{)}^2-(\log 4{)}^2\right]\lambda$
$\therefore \lambda =-1$
$\therefore \log x=-\log 3=\log \frac{1}{3}$
$\Rightarrow \log y=-\log 4=\log \frac{1}{4}$
$\therefore x=\frac{1}{3},y=\frac{1}{4}$