## Calculus with Applications (10th Edition)

$$2 ^{x^{2}-4 x} =\left(\frac{1}{16}\right)^{x-4}$$ The solution of the given equation is $$x =-4 \quad \text { or } \quad x=4$$
$$2 ^{x^{2}-4 x} =\left(\frac{1}{16}\right)^{x-4}$$ Since the bases must be the same, write $\frac{1}{16}$ as $2^{-4 }$ giving \begin{aligned} 2 ^{x^{2}-4 x} &=\left(\frac{1}{16}\right)^{x-4} \\ 2 ^{x^{2}-4 x} &=\left(2^{-4}\right)^{x-4} \\ 2 ^{x^{2}-4 x}&=2^{-4 x+16} \\ x^{2}-4 x &=-4 x+16 \\ x^{2}-16 &=0 \\(x+4)(x-4) &=0 \\ x &=-4 \quad \text { or } \quad x=4 \end{aligned} So, the solution of the given equation is $$x =-4 \quad \text { or } \quad x=4$$