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How do you solve the following equation, preferably algebraically.

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

Feng
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Subhajit Datta
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    Well, we see that $3^x$ grows quite much faster than $4x+1$, so we expect this to have a finite and quite small number of solutions. Did you find any solutions? One solution can be immediately found by considering both sides of the equation in $\mod{3}$. – Matti P. Sep 02 '19 at 05:43
  • You could argue that $3^x$ is "concave up", and $4x+1$ is linear, so there are at most $2$ solutions. Since you can find $2$ integer solutions fairly easily, you are done. – Cheerful Parsnip Sep 02 '19 at 06:01

1 Answers1

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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.

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).

Arthur
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  • I was trying to use the Lambert W function, but couldn't use it. Can you explain the -1 term that you used in LambertW[-1,...], and how that forces the computation away form x = 0? – Subhajit Datta Sep 03 '19 at 06:18
  • @SubhajitDatta That's specific for WolframAlpha. The Lambert $W$ function (which WA actually prefers to call ProductLog) is multivalued, and that argument tells WA which branch to pick. The "standard" branch is $0$, and the other real-valued branch is $-1$. If you're using a different tool, then they might have a different way to distinguish these. – Arthur Sep 03 '19 at 06:22