Question

$(0.5{)}^{3+p}.(0.25{)}^{-8q}=1$ and $9^{q+2}=(\sqrt{3}{)}^{p+1}$. Then $6.(p+q)=$

Answer 1

$(0.5{)}^{3+p}.(0.25{)}^{-8q}=1\Rightarrow 2^{-(p+3)}{2}^{(-2)(-8q)}=2^0$

$\Rightarrow 2^{16q-p-3}=2^0\Rightarrow p+3=16q...\left(1\right)$

Also,

$(9{)}^{q+2}=(\sqrt{3}{)}^{p+1}\Rightarrow 3^{2q+4}=3^{2q+4}=3^{\frac{p+1}{2}}\Rightarrow p=4q+7...\left(2\right)$

Solving (1) & (2)

$\Rightarrow 16q-3=4q+7\Rightarrow q=5/6$

$\Rightarrow p=62/6\therefore p+q=\frac{67}{6}$