Question

Given: ${\log }_{10}x=2a$ and ${\log }_{10}y=\frac{b}{2}$

If ${\log }_{10}P=3a-2b$, then $P$ in terms of $x$ and $y$ is $P=\frac{x^{\frac{3}{2}}}{y^k}$.

then value of $k$ is

Answer 1

$\frac{b}{2}={\log }_{10}y$
Multiplying by 4
$2b=4{\log }_{10}y$ and $2a={\log }_{10}x$

$\therefore \frac{3}{2}\times 2a=\frac{3}{2}{\log }_{10}x$

$3a={\left({\log }_{10}x\right)}^{\frac{3}{2}}$
Solving for log P
$\log P=3a-2b={({\log }_{10}x)}^{\frac{3}{2}}-{\log }_{10}{y}^4$

$\log P=\log \frac{x^{\frac{3}{2}}}{y^4}$

$So,P=\frac{x^{\frac{3}{2}}}{y^4}$

The value of k is 4