Algebraic solution for $4x + 1 = 3^x$

Question

How do you solve the following equation, preferably algebraically.

$$4 x + 1 = 3^x$$

Answer 1

You can't really do it in any nice, constructive way. The $4x$ and the $3^x$ really get in eachother's way.

What you can do it show that there are at most two solutions (at least you can with calculus), and figure out bounds for where those solutions could possibly be. And as you check where the solutions can and can't be, you stumble across the fact that $x=0$ and $x=2$ are solutions.

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Technically, there is something you can do, but it's not pretty. Let's rearrange the equation a bit: $$ 4x+1 = 3^x\\ (4x+1)3^{-x} = 1\\ -\frac14(4x+1)3^{-x} = -\frac14\\ \left(-x-\frac14\right)3^{-x} = -\frac14\\ \left(-x-\frac14\right)3^{-x}\cdot 3^{-1/4} = -\frac14\cdot 3^{-1/4}\\ \left(-x-\frac14\right)3^{-x-1/4} = -\frac14\cdot 3^{-1/4}\\ $$ Looking at this, we see that we have something of the form $s\cdot 3^s = t\cdot 3^t$. One way these can be equal is, of course, if $s = t$, which is to say $-x-\frac14 = -\frac14$. This gives $x = 0$ (just finding this one solution is really lucky; had it been $4x+2$, we wouldn't have had a chance).

The other solution isn't that easy to get a hold of. The best way I know is to use logarithms to rewrite the exponents to have base $e$ instead, multiply on both sides by $\ln 3$. This gives $$ \ln(3)\left(-x-\frac14\right)e^{\ln(3)(-x-1/4)} = -\frac{\ln(3)}4\cdot 3^{-\ln(3)/4} $$ (note that we now have $s\cdot e^s = t\cdot e^t$). We can then apply the so-called Lambert $W$ function, which is made to disentangle $s\cdot e^s$, to both sides of the equation. This way we find the second solution (the -1 in LambertW[-1, is to force the computation away from the $x = 0$ solution).