Find the real value(s) solution of the equation $3^x = 3(x+6)$

Question

I tried using Lambert W function but I got stuck while trying to set the question in the form $xe^x =y$.

Answer 1

To solve using generalised logarithm and Lambert-$W$ function, check the image in https://drive.google.com/file/d/1tw9wiR-q79IVCAQ7fKuXhHQrRtEzwuiD/view?usp=drivesdk

Answer 2

Set $y=x+6$. Then we have $$3^{-6}3^{x+6}= 3(x+6)$$ $$\frac1{729}3^y=3y$$ and you should be able to get to the $ze^z=y$ form now.

Answer 3

By inspection, $x=3$ is a solution.

As you have the intersection between an exponential, a convex function, and a straight line, and $x=3$ is not a double root, there must be another root.

Noting that

$$3^{-6}\approx3(-6+6)=0$$ we can start Newton iterations and get

$$6,\\-5.9995425228212,\\-5.9995425227634,\\-5.9995425227634,\\\cdots$$

Answer 4

$\require{begingroup} \begingroup$

$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

Consider \begin{align} 3^x &= 3(x+6) \tag{1}\label{1} \end{align}
as a special case of

\begin{align} a^x&=b\,x+c ,\quad a=3,b=3,c=18 \tag{2}\label{2} \end{align}

and transform it as follows: \begin{align} \exp(\ln(a)\,x)&=b\,x+c ,\\ \exp(\tfrac 1b\,\ln(a)\,(b\,x))&=b\,x+c; ,\\ \exp(\tfrac 1b\,\ln(a)\,(b\,x+c-c))&=b\,x+c; ,\\ \exp(\tfrac 1b\,\ln(a)\,(b\,x+c)) \,\exp(-\tfrac cb\,\ln(a))&=b\,x+c; ,\\ \exp(-\tfrac cb\,\ln(a))&=(b\,x+c)\,\exp(-\tfrac 1b\,\ln(a)\,(b\,x+c)) ,\\ a^{-c/b}&=(b\,x+c)\,\exp(-\tfrac 1b\,\ln(a)\,(b\,x+c)) ,\\ -\tfrac 1b\,\ln(a)\,a^{-c/b} &= -\tfrac 1b\,\ln(a)\,(b\,x+c)\,\exp(-\tfrac 1b\,\ln(a)\,(b\,x+c)) . \end{align}

At this point we have the right-hand side in a proper form $y\e^y$ and can apply the Lambert $\W$ function:

\begin{align} \W\Big(-\tfrac 1b\,\ln(a)\,a^{-c/b}\Big) &= \W\Big(-\tfrac 1b\,\ln(a)\,(b\,x+c)\,\exp(-\tfrac 1b\,\ln(a)\,(b\,x+c))\Big) ,\\ -\tfrac 1b\,\ln(a)\,(b\,x+c) &= \W\Big(-\tfrac 1b\,\ln(a)\,a^{-c/b}\Big) ,\\ x&=-\frac 1{\ln(a)}\,\W(-\tfrac 1b\,\ln(a)\,a^{-c/b})-\frac cb . \end{align}

Substitution of the values from \eqref{2} gives

\begin{align} x&=-\frac 1{\ln 3}\,\W(-\tfrac 1{2187}\ln 3)-6 . \end{align}

The argument of $\W$ is

\begin{align} -\tfrac 1{2187}\ln 3 &\approx -0.0005 \end{align}

fits inside the interval $(-\tfrac1\e,0)$, where the Lambert $\W$ function has two real branches, $\Wp$ and $\Wm$, so we must have two real solutions to \eqref{1}:

\begin{align} x_0&=-\frac 1{\ln 3}\,\Wp(-\tfrac 1{2187}\ln 3)-6 \approx -0.5025901139 ,\\ x_1&=-\frac 1{\ln 3}\,\Wm(-\tfrac 1{2187}\ln 3)-6 =3. \end{align}

In the second case the value of $x_1=3$ is exact integer, since the argument of $\W$ can also be represented in a form $y\,\e^y$:

\begin{align} -\tfrac 1{2187}\ln 3 &= -\tfrac 1{3^7}\ln 3 = -\tfrac 9{3^9}\ln 3 = \tfrac 1{3^9}\,\ln(\tfrac 1{3^9}) \\ &= \ln(\tfrac 1{3^9})\,\exp\Big(\ln(\tfrac 1{3^9})\Big) = -9\,\ln3\,\exp(-9\,\ln3) . \end{align}

The numeric value \begin{align} -9\,\ln3&\approx -9.88751 \end{align}

is less than $-1$, and belongs to the range of $\Wm$, so in this case we have \begin{align} x_1& =-\frac 1{\ln 3}\,\Wm(-9\,\ln3\,\exp(-9\,\ln3))-6 =-\frac 1{\ln 3}\,(-9\,\ln3)-6 =3. \end{align}

$\endgroup$

Answer 5

Since you have the answers to your question, let us make the problem more funny searching the solution of $$3^x=3x+a$$ without Lambert function.

Consider that you look for the zero's of function $$f(x)=3^x-3x-a$$ the derivatives of which being $$f'(x)=3^x \log (3)-3\qquad \text{and} \qquad f''(x)=3^x \log ^2(3)\quad > 0\quad \forall x$$

The first derivative cancels at $$x_{m}=\frac{1}{\log (3)}\log \left(\frac{3}{\log (3)}\right)$$ and $$f(x_{m})=-a+\frac{3 }{\log (3)}\log \left(\frac{e\log (3)}{3}\right)$$ and there will be two solutions if $f(x_{m})<0$ that is to say $a >- 0.0125$. So, for any positive value of $a$, we shall have two roots such that $x_1< x_{m}$ and $x_2 > x_{m}$.

Planning to use Newton method, we need starting guesses. We can obtain them expanding $f(x)$ as a Taylor series built at $x=x_{m}$ and solving the simple quadratic equation in $(x- x_{m})$ get the estimates $$x_1^{(0)}=x_{m}-\sqrt{-2 \frac{f(x_{m})}{f''(x_{m})}}\qquad \text{and} \qquad x_2^{(0)}=x_{m}+\sqrt{-2 \frac{f(x_{m})}{f''(x_{m})}}$$ Applied to your case where $a=18$, this would give $x_1^{(0)}=-2.39$ and $x_2^{(0)}=4.22$. This is not fantastic but sufficient. For you case, starting with these crude estimates, the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & -2.39173 \\ 1 & -6.07333 \\ 2 & -5.99954 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & 4.22051 \\ 1 & 3.56331 \\ 2 & 3.15154 \\ 3 & 3.01306 \\ 4 & 3.00010 \\ 5 & 3.00000 \end{array} \right)$$

All of that can very easily done with Excel or even a pocket calculator. Just for the fun of it, try any other value of $a$.