Question

Logaritmo
Só faça se você entende bem

Answer 1

$\sqrt{9^{p+1}} = 3^{\sqrt{2}}\\ \\ (\sqrt{9^{p+1}})^2 = (3^{\sqrt{2}})^2\\ \\ 9^{p+1} = 3^{2\sqrt{2}}\\ \\ 3^{2(p+1)} = 3^{2\sqrt{2}}\\ \\ 2(p+1) = 2\sqrt{2}\\ \\ p +1 = \sqrt{2}\\ \\ \boxed{p=\sqrt{2}-1}$

$log_2\ (q-1) = \frac{1}{2}\\ \\ 2^{\frac{1}{2}} = q-1\\ \\ \sqrt{2} = q-1\\ \\ \boxed{q=\sqrt{2}+1}$

Sabemos que:

$\boxed{(p+q)^2 = p^2+2pq+q^2}$

Então:

$(p+q)^2 = (\sqrt{2}-1+\sqrt{2}+1)^2\\ \\ (p+q)^2 = (2\sqrt{2})^2\\ \\ \large\boxed{\therefore\ (p+q)^2 = 8}$