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

Question

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?

Answer 1

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 $y^2-y-3$.

Answer 2

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

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

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