Exponent challenge 144

Step 1: Solve for $n.$

$\begin{aligned} (n+3)^2 & = 16 \\n + 3 & = -\sqrt{16} ~~~~~~ [ n<1] \\n & = -4 - 3 \\ \implies n& = -7. \\ \end{aligned}$

Step 2: Solve for $x.$

$\begin{aligned} \dfrac1{x^{n-1}} & = 3 \\ \dfrac 1{x^{-7-1}} & = 3 \\ \dfrac 1{x^{-8}}& = 3 \\ \dfrac13 & = \dfrac 1{x^8} \\ x^8 & = 3 \\ \implies x & = \pm3^{\frac18}. \\ \end{aligned}$

Step 3: Using $x = \pm 3^{\frac 18},$ find $x^{32} +7$ .

$x^{32} + 7 \\ = (\pm3^{\frac 18})^{32} + 7 \\ = (\pm3)^{\frac 18 \times 32} + 7 \\ = (\pm3)^4 + 7 \\ = 81 + 7 \\ = ~ \boxed{88}.$

$(n+3)^2 = 16 \implies \begin{cases} n+3 = 4 \\ n+3 = -4 \end{cases} \implies \begin{cases} \cancel{n = 1} \quad [n<1] \\ \boxed{n = -7} \end{cases}$

$\begin{aligned} & \implies \frac{1}{x^{(-7)-1}} = 3 \\ & \implies \frac{1}{x^{-8}} = 3 \\ & \implies x^8 = 3 \\ & \implies x^{32} = 3^4 \\ & \implies x^{32} + (7) = 3^4 + (7) \\ & \implies x^{32} + 7 = \boxed{88} \end{aligned}$