How to solve $x + 3^{x} = 4$ using Lambert W Function.

Question

As stated in the title I am trying to solve the equation $$x + 3^{x} = 4$$ using Lambert W Function and which led me to the result $$x = 4 - \frac{W(3^{4} \ln{3})}{\ln{3}}$$ and driven by the belief that Lambert W Function can't be solved, when I entered only the last term in RHS in WolframAlpha the value of $\frac{W(3^{4} \ln3)}{\ln3}$ turned out to be 3 which matches with the final result of $x = 1$ which can be easily deduced just by looking at the equation long enough. But wasn't able to calculate the value of above expression on my own and that's where I need help because I think if the above expression has exactly the value equalt to 3 there must be some way to solve it. Thus, I am interested to know the way to deduce the value of above term by solving the Lambert W Function.

Thanks for any help.

Answer 1

Answer to your explicit question :

Observe that ,

$$ \begin{align}W\left(3^4\ln 3\right)&=W\left(3^3\cdot 3\ln 3\right)\\ &=W\left(3^3\ln 3^3\right)\\ &=W\left(\color{#c00}{\ln 3^3}\cdot e^{\color{#c00}{\ln 3^3}}\right)\\ &=\ln 3^3=3\ln 3\end{align} $$

Therefore, you obtain that :

$$ \begin{align}x&=4-\frac{W\left(3^4\ln 3\right)}{\ln 3}\\ &=4-\frac {3\ln 3}{\ln 3}\\ &=4-3=1\thinspace .\end{align} $$

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$\rm{Construction :}$

Substituting $\thinspace 3^x=u\thinspace $, you get $\thinspace x=4-u\thinspace $ or $\thinspace x=\dfrac {\ln u}{\ln 3}\thinspace $, which leads to the following :

$$ \begin{align}x+3^{x}&=4\\ \frac{\ln u}{\ln 3}+u&=4\\ \ln u+u\ln 3&=4\ln 3\\ \ln \left(u\cdot e^{u\ln 3}\right)&=4\ln 3\\ u\ln 3\cdot e^{u\ln 3}&=e^{4\ln 3}\cdot \ln 3\\ u\ln 3&=W\left(3^4\ln 3\right)\end{align} $$

Then, the exact value becomes :

$$x=4-\frac{W\left(3^4\ln 3\right)}{\ln 3}\thinspace .$$

Answer 2

$\newcommand{\W}{\mathfrak{W}}$I'm not entirely sure what your question is. Say you want to solve: $$x+a^x=b$$Recall to use $\W$ we would like to have terms of the form $xe^x$. Rearrange to: $$\ln(a)\cdot a^b=\ln(a)(b-x)a^{b-x}$$It follows that: $$\W(\ln(a)\cdot a^b)=\ln(a)(b-x)$$So: $$x=b-\frac{1}{\ln a}\W(\ln(a)\cdot a^b)$$

Are you asking how to recognise that: $$\frac{\W(3^4\ln 3)}{\ln 3}=3$$? In which case, define $a=\frac{1}{\ln 3}\W(3^4\ln 3)$ and note: $$a\ln(3)\cdot e^{a\ln 3}=\W(3^4\ln 3)\cdot e^{\W(3^4\ln 3)}=3^4\ln 3$$So it follows that: $$a\cdot 3^a=3^4$$And from there you just have to solve by inspection that $a=3$. More generally we have the result: $$\frac{\W(x^{x+1}\ln x)}{\ln x}=x$$

Answer 3

Let $y=4-x$

$$3^{4-y}=y\Longrightarrow 3^4=y\cdot 3^y=y\cdot e^{y\ln3}\Longrightarrow 3^4\ln3=(y\ln3)\cdot e^{y\ln3}$$

Use Lambert W-function,

$$y\ln3=W(3^4\ln3)$$

Finally, go back to $x$

$$x=4-\frac{W(3^4\ln3)}{\ln3}$$

Answer 4

$$x＋3^x＝4\implies3^{x＋3^x}＝3^4\implies3^x3^{3^x}＝3^4\\~\\\text{let :}y＝3^x\;\implies\;y3^y＝3^4\implies\;y＝W_3\left({3^4}\right)\implies3^x＝W_3\left({3^4}\right)\implies\;x＝\frac {\ln(W_3\left({3^4}\right))}{\ln(3)}$$ and therfore $$\implies\;x＝\log_3(3^4)-\frac {W(3^4\ln(3))}{\ln(3)}＝4-\frac {W(3^4\ln(3))}{\ln(3)}\\~\\$$ let t:=$\frac {W(3^4\ln(3))}{\ln(3)}$ $$\implies\;t\ln(3)＝W(3^4\ln(3))\implies\;e^{t\ln(3)}＝e^{W(3^4\ln(3))}\\~\\\implies\;t\ln(3)e^{t\ln(3)}＝W(3^4\ln(3))e^{W(3^4\ln(3))}＝3^4\ln(3)\\~\\\implies\;t\ln(3)e^{t\ln(3)}＝3^4\ln(3)\implies\;t＝3$$ Finally: $$x＝4-\frac{W(3^4\ln(3))}{\ln(3)}＝4-3＝1\implies\;x＝1$$