Solving Equation using Lambert W function

Question

Would somebody be able to talk me through solving the following logarithmic inequality for $x$ using the Lambert W function?

$$x\ln(2) - 61\ln(x+2) > 200$$

I already know this: Wikipedia: Lambert W function.

I also know how inequalities are solved by solving equations.

And I know that equations of the form $a+bx+c\ln(x)=d$, wherein $a,b,c,d$ are constants, can be solved using the Lambert W function - but I don't know how this can be done in the general cases. And now we have additionally the both terms $x$ and $\ln(x+2)$.

Answer 1

$x\ln 2 - 61 \ln (x+2) > 200$

$\frac {2^x}{(x+2)^{61}} > e^{200}$

$\frac 14 2^{x+2}(x+2)^{-61} > e^{200}$

$2^{x+2}(x+2)^{-61} > 4e^{200}$

$(\frac 1{2^{\frac 1{61}}})^{x+2}(x+2) < \frac 1{(4e^{200})^{\frac 1{61}}}$

$0> e^{(x+2)\ln (\frac 1{2^{\frac 1{61}}})}[(x+2)\ln (\frac 1{2^{\frac 1{61}}})]> \ln (\frac 1{2^{\frac 1{61}}})\frac 1{(4e^{200})^{\frac 1{61}}}$

To solve $-\frac 1e< -\frac 12 e^{-261} < e^{-x-2}(-x-2)< 0$ we use the decreasing $W_{-1}$ function and

$(x+2)\ln (\frac 1{2^{\frac 1{61}}}) < W_{-1}(\ln (\frac 1{2^{\frac 1{61}}})\frac 1{(4e^{200})^{\frac 1{61}}})$

$x > -2-\frac 1{61}\ln 2 W_1(-61\ln 24^{-\frac 1{61}}e^{-\frac {200}{61}})$

Answer 2

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

Consider $f(x)=x\ln(2) - 61\ln(x+2) -200$ for $x>-2$. It has one minimum at $x=-2+\tfrac{61}{\ln2}$, so in order to solve $f(x)>0$, you just need to find two zeros of $f(x)$ in terms of the Lambert W function, which can be done as follows:

\begin{align} x\ln(2) - 61\ln(x+2) -200&=0 ,\\ (x+2)\ln2-2\ln2 - 61\ln(x+2)-200&=0 ,\\ (x+2)\ln2-61\ln(x+2)&=200+2\ln2 ,\\ -\tfrac{\ln2}{61}(x+2)+\ln(x+2) &= -\tfrac{200}{61}-\tfrac2{61}\ln2 ,\\ \ln(\exp(-\tfrac{\ln2}{61}(x+2)))+\ln(x+2) &=-\tfrac{200}{61}-\tfrac2{61}\ln2 ,\\ \ln((x+2)\exp(-\tfrac{\ln2}{61}(x+2))) &=-\tfrac{200}{61}-\tfrac2{61}\ln2 ,\\ (x+2)\exp(-\tfrac{\ln2}{61}(x+2)) &=\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2) ,\\ -\tfrac{\ln2}{61}(x+2)\exp(-\tfrac{\ln2}{61}(x+2)) &=-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2) ,\\ \W\left(-\tfrac{\ln2}{61}(x+2)\exp(-\tfrac{\ln2}{61}(x+2)) \right) &=\W\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right) ,\\ -\tfrac{\ln2}{61}(x+2) &=\W\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right) ,\\ x &=-2-\tfrac{61}{\ln2}\W\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right) . \end{align}

Now, as we have the solution of $f(x)=0$ in terms of the Lambert W function, we need to check its argument \begin{align} z&=-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2) \approx -0.0004185. \end{align}

As we can see, $z\in (-\tfrac1\e,0)$, so indeed we have two solutions

\begin{align} x_0 &= -2-\tfrac{61}{\ln2}\Wp\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right) \approx -1.963153548947 ,\\ \text{and }\quad x_1 &=-2-\tfrac{61}{\ln2}\Wm\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right) \approx 885.9981973778 . \end{align}

Thus, the answer is

\begin{align} f(x)&>0 \quad \text{for }\quad x\in (-2,x_0) \cup x\in(x_1,\infty) . \end{align}

$\endgroup$