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? 
 A: 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.
A: 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.
A: $$3^x=9x \Rightarrow 1 =\frac {9x}{3^x} \Rightarrow 1= 9x \cdot e ^{-x\ln 3}\Rightarrow \frac{1}{9}=x \cdot e ^{-x\ln 3} \Rightarrow$$
$$\Rightarrow \frac{-\ln 3}{9}=(-x \ln 3)\cdot e^{-x\ln 3}\Rightarrow W\left(\frac{-\ln 3}{9}\right)=-x \ln 3\Rightarrow$$
$$x= \frac{-W\left(\frac{-\ln 3}{9}\right)}{\ln 3}$$
where W is Lambert W function .
A: 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.
A: I think the only way of manipulating this jargon is by splitting the equation .
$3^x=9x$
$y=3^x$     and    $  y=9x$
when plotted, you find that the two equations intersect at $x=3$
therefore,$x=3$
A: I am going to give three methods of solving this problem and they are:
$1$. Differentiation -                                                                               It can be broken down into a simpler form by finding  the derivative of both sides.
$\frac{d(3^x)}{dx} = \frac{d(9x)}{dx}$
$x(3^x - 1) = 9$
$x = 9$
$3^{x - 1}=9$
$3^{x - 1} = 3^2$
Equating the exponents,
$x - 1 = 2$
$x = 3$
Thus $x = 3 \ or \ 9$.
But since $9$ does not satisfy the equation.
$x = 3$.
$2$. Graphical approach -
This can be done by plotting a graph of $y=9x$ and $y=3^x$ and the $x$ component of the intersection is going to be $3$.
Thus $x=3$.
3. Value analyzing - This done by replacing the variable with figures starting from zero to see which one satisfies the equation and you will find out the equation is true when $x$ is $3$.
Thus $x=3$.
