Find x when the function equals 0 I must solve for x for this function. 
$e^x-20x=0$
I'm not sure what to do here. I've tried this so far but it makes no sense:
$$\begin{align*}
e^x&=20x\\
x\ln e&=\ln20+\ln x\\
\frac{x}{\ln20}&=\ln x
\end{align*}$$
I tried this as well but I'm not sure if this is right either:
$$\begin{align*}
e^x&=20x\\
\ln e^x&=\ln20x\\
x&=\ln20+\ln x
\end{align*}$$
I've got more confidence in the second try, although I don't know how to solve for an actual number. 
The answer is a decimal. 
 A: This equation cannot be solved in terms of elementary functions. You can either use numeric methods like Newton's, or solve in terms of Lambert W function:
\begin{align*}
e^x - 20x &= 0 \\
20x &= e^x \\
xe^{-x} &= \frac{1}{20} \\
-x e^{-x} &= -\frac{1}{20} \\
-x &= W\left(-\frac{1}{20}\right) \\
x &= -W\left(-\frac{1}{20}\right)
\end{align*}
A: You will not get an exact answer.  Use a numerical method such as the Newton Method.  It will converge quickly to a very close approximate solution.
A: You can try to solve this iteratively by isolating $x$ and replacing repeatedly (you can even do this in a spreadsheet). Though this is not guaranteed to work. In this case, it works and the possible ways gives you the two solutions:
$$e^x - 20x = 0   \Rightarrow x = \frac{\exp(x)}{20}$$
Then we start with (say $x_0=1$) and make $x_1 = \frac{\exp({x_0})}{20}$, $x_2 = \frac{\exp({x_1})}{20}$ ... and you quickly get one solution $x\approx 0.0527$
We can also try 
$$e^x - 20x = 0   \Rightarrow x = \log(20 x)$$
and do the same: we get $x\approx 4.499755$
But don't assume this method will always work so nicely.
A: The function $f(x) = e^x - 20x$ is obviously positive when $x$ is negative and $f(0)=1$.
The derivative $f'(x) = e^x - 20$ has a zero in $x = \ln(20) \approx 2,996$. The second derivative is always positive so the point is a minimum.
In $x = \ln(20)$ the function takes the value 
$$
f(\ln(20)) = 20 - 20\ln(20) = 20(1-\ln(20)) = -20(\ln(2)-1)\approx -39,915
$$
So the equation has two distinct solutions. The first one is before $1$. In fact $e^1 = e$ is smaller than 20. The second one is between 4 and 5.
You cannot, with simple alegebraic manipulation, arrive to an expression of the solutions. But now that you know the number of solutions and where they are you can find them using numerical methods.
