Question

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

Answer 1

We have,

${\left(0.5\right)}^{3+p}\times {\left(0.25\right)}^{-3q}=1$

$9^{q+2}={\left(\sqrt{3}\right)}^{p+1}$

Therefore,

${\left(0.5\right)}^{3+p}\times {\left({\left(0.5\right)}^2\right)}^{-3q}=1$

${\left(0.5\right)}^{3+p}\times {\left(0.5\right)}^{-6q}=1$

${\left(0.5\right)}^{3+p}={\left(0.5\right)}^{6q}$

$3+p=6q$ …….. (1)

Similarly,

$9^{q+2}={\left(\sqrt{3}\right)}^{p+1}$

${\left({\left(\sqrt{3}\right)}^4\right)}^{q+2}={\left(\sqrt{3}\right)}^{p+1}$

${\left(\sqrt{3}\right)}^{4q+8}={\left(\sqrt{3}\right)}^{p+1}$

$4q+8=p+1$

$p-4q=7$ ……….. (2)

On solving equations (1) and (2), we get

$p=27,q=5$

Since,

$p+q=27+5=32$

Hence, this is the answer.