The pressure p and the volume v

We have

$p v^{1.4}=$ constant $=k$ (say)

Taking log on both the sides, we get

$\log \left(p v^{1.4}\right)=\log k$

Differentiating both the sides w.r.t. $x$, we get

$\frac{1}{p} \frac{d p}{d v}+\frac{1.4}{v}=0$

$\Rightarrow \frac{d p}{p}=\frac{-1.4 d v}{v}$

Now, $d p=\frac{d p}{d v} d v=\frac{-1.4 p}{v} d v$

$\Rightarrow \frac{d p}{p} \times 100=-1.4\left(\frac{d v}{v} \times 100\right)=-1.4 \times\left(\frac{-1}{2}\right)=0.7$            $\left[\right.$ Since we are given $\frac{1}{2} \%$ decrease in $\left.v\right]$

Hence, the error in $p$ is $0.7 \%$.