please show workings

(2^n)^n x (2^n)^3 x 4 =1

$\begin{array}{rcll} (2^n)^n \cdot (2^n)^3 \cdot 4 &=&1 \\ (2^n)^n \cdot (2^n)^3 \cdot 2^2 &=& 2^0 \\ 2^{n^2} \cdot 2^{3n} \cdot 2^2 &=& 2^0 \\ 2^{n^2+3n+2} &=& 2^0 \\ n^2+3n+2 &=& 0 \\ \end{array}$

$\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}$

$\begin{array}{rcll} n^2+3n+2 &=& 0 \quad | \quad a=1 \quad b = 3 \quad c = 2 \\ n &=& \dfrac{-3 \pm \sqrt{3^2-4\cdot 1 \cdot 2} }{2\cdot 1} \\ n &=& \dfrac{-3 \pm \sqrt{9-8} }{2} \\ n &=& \dfrac{-3 \pm \sqrt{1} }{2} \\ n &=& \dfrac{-3 \pm 1 }{2} \\ \\ n_1 &=& \dfrac{-3 + 1 }{2} \\ n_1 &=& \dfrac{-2 }{2} \\ \mathbf{n_1} & \mathbf{=} & \mathbf{-1} \\ \\ n_2 &=& \dfrac{-3 - 1 }{2} \\ n_2 &=& \dfrac{-4 }{2} \\ \mathbf{n_2} & \mathbf{=} & \mathbf{-2} \end{array}$

Many thanks for your help.

Solve for n over the real numbers:

2^(3 n+2) (2^n)^n = 1

Take the natural logarithm of both sides and use the identities log(a b) = log(a)+log(b) and log(a^b) = b log(a):

log(2) n^2+log(2) (3 n+2) = 0

Expand out terms of the left hand side:

log(2) n^2+3 log(2) n+2 log(2) = 0

The left hand side factors into a product with three terms:

log(2) (n+1) (n+2) = 0

Divide both sides by log(2):

(n+1) (n+2) = 0

Split into two equations:

n+1 = 0 or n+2 = 0

Subtract 1 from both sides:

n = -1 or n+2 = 0

Subtract 2 from both sides:

Answer: |n = -1 or n = -2