# How do I solve this analytically $3^x=9x$

One of my friends ask me how to solve this equation analytically $3^x=9x$. Looking at it I guess 3 is the answer and I also plot a graph of line $9x$ and the curve $3^x$, they intersect at 3.

But, what I want is to give an analytical solution of the equation. I started

$3^x=9x$

$3^{x-2}=x$

$(x-2)\ln3=\ln x$

How can I continue?

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$$3^x=9x \Rightarrow 1 =\frac {9x}{3^x} \Rightarrow 1= 9x \cdot e ^{-x\ln 3}\Rightarrow \frac{1}{9}=x \cdot e ^{-x\ln 3} \Rightarrow$$

$$\Rightarrow \frac{-\ln 3}{9}=(-x \ln 3)\cdot e^{-x\ln 3}\Rightarrow W\left(\frac{-\ln 3}{9}\right)=-x \ln 3\Rightarrow$$

$$x= \frac{-W\left(\frac{-\ln 3}{9}\right)}{\ln 3}$$

where W is Lambert W function .

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There are two roots! – Evgeny Savinov Jan 22 '12 at 13:01
@SavinovEvgeny: the Lambert W function has multiple branches which amount to the different solutions (there are two real solutions and some more in the complex plane). – Fabian Jan 22 '12 at 13:37
Yes, there are two real roots – Manoj Pandey Mar 27 '13 at 12:41

This equation has no analytical solution, you can find roots only numerically. But one root is obvious. And we can prove that there is no root which is more than 3 by taking the derivative of $y=((x-2)\ln(3))-(\ln(x))$ that gives $y'=\ln(3)-\frac{1}{x}>0$ when $x>3$. There is another solution approximately $0.127869$. Because of monotonicity and continuity of $y$ on $[0,\frac{1}{\ln3}]$ there is only one root there, analogically on $(\frac{1}{\ln3},\infty)$. That is only two roots.

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How do you find the other root 0.127869? – Hassan Muhammad Jan 22 '12 at 18:15
You can find the number with any given accuracy. For example by en.wikipedia.org/wiki/Newton%27s_method – Evgeny Savinov Jan 22 '12 at 18:39

It's easy to prove that $3^x \geq 9x$ for all real $x \leq 0$ and $x\geq 3$ with equality at $x=3$. Furthermore, in the interval $(0,3)$ there is exactly one $\alpha$ such that $3^\alpha = 9\alpha$. Put all this together and it's not hard to show that $3^x > 9x$ for $x \in (-\infty,\alpha) \cup (3,\infty)$ and $3^x < 9x$ for $x \in (\alpha,3)$.

There is no way of algebraically manipulating the equation $3^x = 9x$ around to get this(*) - you have to use calculus (or something to that effect).

(*): Unless you don't mind using facts about the Lambert W function, and that amounts to the same as using calculus; essentially.

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If $x \ge 1$, $3^x$ is not always greater than or equal to $9x$. Suppose $x=1$, do you get it? – Hassan Muhammad Jan 22 '12 at 18:18
Yes, $3^2=9, 9*2=18$ $9<18$ it's false – Evgeny Savinov Jan 22 '12 at 18:33
@HassanMuhammad: That was a typo. I fixed it now. – kahen Jan 22 '12 at 18:37

To find "the other root" it often helps to express the original problem as deviation from that root which is already known. So if we "see" one root is $\small x=3$ I'd proceed $$\small 3^x = 9x \to 3^{3+d} = 9(3+d)$$ where $\small d=0$ refers to the already known solution. Then the equation often can be much simplified to exhibit a possible interval for another solution more visibly.

$\qquad \small \begin{eqnarray} 3^{3+d} &=& 9(3+d) \\ 27 \cdot 3^d &=& 27+9d \\ 3^d &=& 1+d/3 \\ \end{eqnarray}$
and search for solutions in $\small d \ne 0$ If we want to avoid calculus here, but know the expression for the exponential series, we might introduce the symbol $\small \lambda = \ln(3)$ and write
$\qquad \small \begin{eqnarray} 1+ \lambda d + (\lambda d)^2/2! + \ldots &=& 1 + d/3 \\ \lambda d + (\lambda d)^2/2! + \ldots &=& d/3 \\ \lambda + \lambda^2 d/2! + \lambda^3 d^2/3! + \ldots &=& 1/3 \\ d( \lambda^2 /2! + \lambda^3 d/3! + \ldots ) &=& 1/3 - \lambda & \lt & 0\\ \end{eqnarray}$

so d must be negative. Then we can write using -c=d and c positive:

$\qquad \small \begin{eqnarray} 3^{-c} &=& 1-c/3 \\ 1/3^c + c/3 &=& 1 \\ \end{eqnarray}$

and have an initial guess ($\small 2 \lt c \lt 3$) for the interval for the second possible c ($\small c \ne 0$) resp the second root $\small d = -c \ne 0 \to 0 \lt x \lt 1$ and might apply Newton's iteration for approximation of the actual root. Also we see immediately that there is no further real (non-complex) root.

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