# 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;

-
Those are the roots of the quadratic equation in $m$. Remember that $m = 4^x$. – Tyler Mar 26 '12 at 1:22
but it doesn't make sense, what would I do for $4^{5±151i√2}$ ? – aajjbb Mar 26 '12 at 1:40
This is very puzzling. – 000 Mar 26 '12 at 1:56
1,3,8,16, or 20 are not solutions of the original equation. – deinst Mar 26 '12 at 1:58
I venture to guess that there is a mistake in the book, or that you have misunderstood and/or miscopied the problem. Maybe if you tell us what book and what page, someone will be able to have a look. – Gerry Myerson Mar 26 '12 at 2:19

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 $k = 1$.

-

\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.

-
If you wanted to allow complex solutions, you'd also have to take into account the multivaluedness of the complex log, so there would be infinitely many solutions, not just two. – Robert Israel Mar 26 '12 at 2:41
@RobertIsrael Yes, that was what I was thinking. However, I do not have a scholarly understanding of those things--so I felt unsure in speaking about them. Thank you for your enlightenment. – 000 Mar 26 '12 at 2:46
Oh, sorry, I the options I typed, was for the sum of the roots, but even summing the two roots, it's does not make sense too. – aajjbb Mar 26 '12 at 12:41
but Robert Israel, I didn't got the $b$ in your solution, for what this stands for ? – aajjbb Mar 26 '12 at 12:44
Since the original equation doesn't work, I'm suggesting one of the numbers in it had a copying error. For no particular reason, I'm supposing it was the last $4$ that should be something else. $b$ stands for that something else. – Robert Israel Mar 26 '12 at 17:57