How to solve $x + 3^x = 4$ analytically?

Question

I tried to solve the problem of $x + 3^x = 4$. I know that we can find it that $x = 1$ intuitively. I just want to know how to solve it using mathematic formula.

I have learned a little bit about Lambert W function. I've tried to rewrite the formula to become: \begin{align} 1 &= (4-x) \, e ^{-x \ln(3)} \\ 1 &= 4 \, e ^{-x \ln(3)} - x \, e ^{-x \ln(3)} \\ 1 &- 4 \, e ^{-x \ln(3)} = -x \, e ^{-x \ln(3)} \\ \ln(3) & - 4 \, \ln(3) \, e ^{-x \ln(3)} = -x \ln(3) \, e ^{-x \, \ln(3)} \end{align}

Until that last part, I got confused how to combine the $x$ variables at both sides. That's why I got stuck. Any idea how to combine it? Or is there any part of my solution need to be corrected? Thank you.

Answer 1

By the methods started by the proposer one can show that the equation in question is part of a more general set given by $x + a^{x} = b$ which has the soltion $$x = b - \frac{W_{0}(a^{b} \, \ln(a))}{\ln(a)}.$$

The proof of which can be seen by use of $a^{x} = e^{x \, \ln(a)}$ and $x \, e^{x} = t$ has the solution $t = W(x)$, where $W(z)$ is the Lambert W-function, where $W_{0}(x)$ is defined as th ereal solution for $x \geq -1/e$, and is: \begin{align} x + a^{x} &= b \\ (x - b) &= - a^x = - e^{(x-b) \, \ln(a) + b \, \ln(a)} \\ -(x - b) \, \ln(a) \, e^{- (x-b) \, \ln(a)} &= a^{b} \, \ln(a) \\ - (x-b) \, \ln(a) &= W_{0}(a^{b} \, \ln(a)) \\ x &= b - \frac{W_{0}(a^{b} \, \ln(a))}{\ln(a)}. \end{align}

For this particular problem one can use the property $W(x \, \ln(x)) = \ln(x)$ in such a way that $W(3^{4} \, \ln(3)) = W(27 \, \ln(27)) = \ln(27) = 3 \, \ln(3)$. For the equation $x + 3^{x} = 4$, which from the general solution is $a=3$ and $b=4$ yields \begin{align} x &= 4 - \frac{W_{0}(81 \, \ln(3))}{\ln(3)} \\ &= 4 - \frac{3 \, \ln(3)}{\ln(3)} \\ &= 1. \end{align}

Answer 2

To get a solution using the Lambert $W$ function which has $z=W(ze^z)$:

$$x+3^x=4$$

$$x-4 = -81\cdot3^{x-4}$$

$$\log_e(3)(x-4) = -81\log_e(3)\cdot e^{\log_e(3)(x-4)}$$

$$-\log_e(3)(x-4)\cdot e^{-\log_e(3)(x-4)}= 81\log_e(3)$$

$$-\log_e(3)(x-4) = W\left(81\log_e(3)\right)$$

$$x=4 - \dfrac{W\left(81\log_e(3)\right)}{\log_e(3)}$$

Answer 3

To hell with the formulas. As the function $x+3^x$ is increasing the solution $x=1$ is unique.

Answer 4

I am assuming we are dealing with the Real domain i.e. $ \{x|x \in \mathbb{R} \} $

Rearrange equation $ x + 3^x = 4$ so that there is a zero on one side: $$ x + 3^x - 4 = 0 $$ Let $ f(x) = x + 3^x - 4 = 0 $

Find potential turning points by solving for the derivative $ \frac{d}{dx} [f(x)] $ and then solving for $\frac{d}{dx} [f(x)] = 0 $

Apply addition rule: $ \frac{d}{dx} [x + 3^x - 4] = \frac{d}{dx}[x] + \frac{d}{dx} [3^x] - \frac{d}{dx} [4] $

Use properties $ \frac{d}{dx}[kx] = k \: and \: \frac{d}{dx}[c] = 0 $

$ = 1 + \frac{d}{dx} [3^x] - 0 $

$ = 1 + \frac{d}{dx} [e^{\ln(3)*x}]$ , using the fact that $ \frac{d}{dx} [e^x] = e^x $ and applying the chain rule $\frac{d}{dx}[f(g(x))] = f'g(x) * g'(x) $

$ 1 + \frac{d}{dx}[e^{\ln(3)*x}] = 1 + e^{\ln(3)*x} * \ln(3) $

$ = 1 + 3^x * \ln(3) $

Now solve for $ 1 + 3^x * ln(3) = 0 $ :

$ 3^x * \ln(3) = -1 $

$ 3^x = -\frac{1}{\ln(3)} $ , negative number, no solutions in the real domain thus there are no stationary points and no turning points.

As a consequence of Rolle's Theorem: $ \text{#Roots} \leq \text{#turningPoints} + 1 $

Thus, $ \text{max#OfRoots} = 0 + 1 = 1 $

Therefore, the only solution in the real domain is $ x = 1$