# Solve for $x$: Equation quadratic in $2^x$—getting two different answers?

The equation is:

$4^x = 2^x +3$

So...

$$4^x = 2^x + 3 \implies (2^x)^2 - 2^x - 3 = 0$$ Which factors to $(2^x - 3)(2^x + 1)$. We can ignore the second factor because the range of $\log{x}$ is strictly greater than zero. Therefore:

$$2^x - 3 = 0 \implies x = \log_2{(3)} \approx 1.585$$

But if we solve this with the quadratic equation, we get:

$$2^x = \frac{1 \pm \sqrt{1 - (-12)}}{2}$$

Again, we can ignore the negative, so:

$$x = \log_2{\left(\frac{1 + \sqrt{13}}{2}\right)} \approx 1.203$$

Usually, by the time I get to this point in writing my question, I find the obvious mistake, which is often an arithmetical one (and I end up deleting my question). But I'm not seeing anything blatantly wrong here. Why am I getting two different answers for two seemingly correct ways of solving the above equation?

• $(2^x - 3)(2^x + 1)=2^{2x}-2^{x\color{red}{+1}}-3$ – dxiv Jun 23 '17 at 0:02
• @dxiv Where does that $x+1$ come in? $2^x \cdot 1 = 2^x$ and $-3 \cdot 2^x = -3(2^x)$, so shouldn't $2^x - 3(2^x) = -2(2^x)$ ? – AmagicalFishy Jun 23 '17 at 0:06
• Yes, but can you see why that is the same? – Improve Jun 23 '17 at 0:07
• Fishy, could you please do these: (I) find $(u-3)(u+1)$ and (II) factor $w^2 - w - 3$ – Will Jagy Jun 23 '17 at 0:11
• Fishy, please try the thing with a substitute variable, as I suggested. Reduces errors – Will Jagy Jun 23 '17 at 0:13

## 2 Answers

Write $y=2^x$, so the quadratic equation is $y^2-y-3=0$. This does not factor to $(y-3)(y+1)$, because the latter expands to $y^2-3y+y-3=y^2-2y-3$. So your first calculation is incorrect.

Your second calculation is correct, and also demonstrates that there is not a nice factorisation for the quadratic $y^2-y-3$.

• Oooh! Oh, man. I feel dumb—now I see. I've been using the $2$ in $2^x$ as an extra coefficient. :l – AmagicalFishy Jun 23 '17 at 0:08
• @AmagicalFishy No need to feel dumb. Every mathematician makes mistakes of this kind every now and then (sometimes in more advanced contexts, but it still makes us feel dumb when it shouldn't). – Ethan Bolker Jun 23 '17 at 0:15
• @AmagicalFishy As Ethan says, this is nothing to worry about: most of us spend a couple of days baffled by a calculation that gives nonsense, only to eventually find that we made a crucial sign error! – Chappers Jun 23 '17 at 0:21

Better to use extra letters $2^x = w$

$$w^2 - w - 3 = 0$$ In this version it is clear that there is no factoring

$$w = \frac{1 \pm \sqrt {13}}{2}$$

• This doesnt really answer the question – Improve Jun 23 '17 at 0:05
• @Improve poor handling of the $2^x$ term seems to be exactly the problem here, and this would not happen with a dummy variable. As in the comments now below the main question. – Will Jagy Jun 23 '17 at 0:09