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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;

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    Feeding it to Wolfram Alpha as http://www.wolframalpha.com/input/?i=plot+16%5Ex%2B44-5*4%5Ex or http://www.wolframalpha.com/input/?i=solve+16%5Ex%2B44-5*4%5Ex%3D0 shows no real roots and no good set of complex roots that would give a simple real sum. So I would support Gerry Myerson's guess that there is an error in the problem.2012-03-26

2 Answers 2

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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 $k = 1$.

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$ \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.

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    @aajjbb, I'm happy for you! I recommend reporting this misprint to the author--I do that whenever I encounter an error in a book.2012-03-27