Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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



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

How can I continue?

share|cite|improve this question
up vote 11 down vote accepted

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

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

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

where W is Lambert W function .

share|cite|improve this answer
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.

share|cite|improve this answer
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 – 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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.