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Solve the equations

Let $P$ be the product of all non-zero solutions.

Find $48 P$ .

Hint

For part (a), we know that $16 = 2^{4}$ , so we can write

\begin{aligned} 2^{x^{2}} & = {(2^{4})}^{(x + 3)} \\ 2^{x^{2}} & = 2^{4 (x + 3)} \end{aligned}

which is true if (and only if)

x^{2} = 4 (x + 3)

Parts (b), (c), (d) can be approached in a similar way.

Solution

Going by the hint, we have
$\begin{aligned} 2^{x^{2}} & = {(2^{4})}^{(x + 3)} \\ 2^{x^{2}} & = 2^{4 (x + 3)} \\ x^{2} & = 4 (x + 3) \\ x^{2} - 4 x - 12 & = 0 \\ (x - 6) (x + 2) & = 0 \\ x & \in {- 2, 6} \end{aligned}$
Again, we aim to get the same base on both sides:
$\begin{aligned} {(3^{2})}^{x^{2}} & = 3^{5 x - 3} \\ 3^{2 x^{2}} & = 3^{5 x - 3} \\ 2 x^{2} & = 5 x - 3 \\ 2 x^{2} - 5 x + 3 & = 0 \\ (2 x - 3) (x - 1) & = 0 \\ x & \in {\frac{3}{2}, 1} \end{aligned}$
We aim to express both sides in base $2$ :
$\begin{aligned} {(2^{3})}^{x^{2}} & = \frac{{(2^{2})}^{4 x}}{2^{4}} \\ 2^{3 x^{2}} & = 2^{8 x - 4} \\ 3 x^{2} & = 8 x - 4 \\ 3 x^{2} - 8 x + 4 & = 0 \\ (x - 2) (3 x - 2) & = 0 \\ x & \in {2, \frac{2}{3}} \end{aligned}$
Don't be fooled by the notation: use the laws of exponents.
$\begin{aligned} 2^{2} x & = 2^{x^{2}} \\ 2 x & = x^{2} \\ x^{2} - 2 x & = 0 \\ x (x - 2) & = 0 \\ x & \in {0, 2} \end{aligned}$