# 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, 2017 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 $2^x - 3(2^x) = -2(2^x)$ ? Jun 23, 2017 at 0:06
• Yes, but can you see why that is the same? Jun 23, 2017 at 0:07
• Fishy, could you please do these: (I) find $(u-3)(u+1)$ and (II) factor $w^2 - w - 3$ Jun 23, 2017 at 0:11
• Fishy, please try the thing with a substitute variable, as I suggested. Reduces errors Jun 23, 2017 at 0:13

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 Jun 23, 2017 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). Jun 23, 2017 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! Jun 23, 2017 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 Jun 23, 2017 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. Jun 23, 2017 at 0:09