Exponential Equation with mistaken result I'm on my math book studying exponential equations, and I got stuck on this Problem:
What is sum of the roots of the equation: 
$$\frac{16^x + 64}{5} = 4^x + 4$$
I decided to changed: $4^x$ by $m$, so I got: $$\frac{m^2 + 64}{5} = m + 4$$
working on it I've got: $m^2 - 5m + 44 = 0$
but solving this equation the roots were: $x = \frac{5 \pm \sqrt{151i}}{2}$ 
which isn't even close from the possible answers: 1, 3, 8, 16 or 20.
What's the mistake ? thanks in advance;
 A: $$
\begin{align}
\frac{16^x+64}{5}&=4^x+4\\
16^x+64&=5\cdot 4^x+20\\
16^x-5\cdot 4^x&=-44\\
y^2-5y+44&=0 \quad \text{for }  y=4^x\\
y&=\frac{1}{2}\left(5\pm i\sqrt{151}\right)\\
4^x&=\frac{1}{2}\left(5\pm i\sqrt{151}\right)\\
x&=\log_{4}\left( \frac{1}{2}\left(5\pm i\sqrt{151}\right)\right)\\
\sum(x)&=\log_{4}\left( \frac{1}{2}\left(5+ i\sqrt{151}\right)\right)+\log_{4}\left( \frac{1}{2}\left(5- i\sqrt{151}\right)\right)\\
&=2\Re\left(x\right)\\
&\approx 2\cdot 1.89209=3.78418
\end{align}
$$
This is the logical answer I arrived at. Is there an error in the problem itself?
It seems that this becomes a very complex problem when it is taken to the complex plane. (I have omitted those answers as a result, since I do not understand them.) Perhaps someone more enlightened than myself would elaborate.
A: Perhaps you copied the equation wrong.  For the equation
$$ \frac{16^x + 64}{5} = 4^x + b$$
(where presumably $x$ is supposed to be real), substituting $m = 4^x$ we get
$$m = \frac{5 \pm \sqrt{20 b - 231}}{2}$$
where we want  both solutions to be real, so $11.55 \le b < 12.8$.  Now
$x_1 + x_2 = k$ where $m_1 m_2 = 4^{x_1 + x_2} = 4^k$.  In this case
$$m_1 m_2 = \frac{5^2 - (20 b - 231)}{4} =  64 - 5 b$$
Thus if $b=12$ you get $4^k = 4$ so $k = 1$.
