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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.

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wolframalpha.com/input/?i=x+%3D+ln%2820%29+%2B+ln%28x%29. You need to do this numerically. –  M.B. Jun 1 '12 at 0:10
    
There is no way to simply "solve for $x$" analytically. You need to use the Lambert W function to express $x$. You can find numerical answers, though, by noting that the function is positive at $0$ and negative at $1$, so it must be zero somewhere in between; then checking at $0.5$, and so on, to approximate the answer to whatever degree of accuracy you want. –  Arturo Magidin Jun 1 '12 at 0:10
    
Equations like this where the unknown variable is in an exponent and down on the "ground level" very rarely admit the kind of solution you are looking for. If this problem came from a pre-calculus class, I would think that you are intended to use an exploratory approach: use a graph or a table to find an approximate decimal value to the solution. Later, in calculus you can learn processes that will give as many decimals as you can handle, but never a solution in terms of the basic functions. –  alex.jordan Jun 1 '12 at 0:12
    
For what it's worth, do you mean $x$ instead of $c$ in line 2? And the third line is not correct; you wanted to subtract that $\ln 20$. –  alex.jordan Jun 1 '12 at 0:14
    
@alex.jordan Yes, and oops =) @ everyone else, Thanks for the help! –  ninja08 Jun 1 '12 at 0:16

4 Answers 4

up vote 3 down vote accepted

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*}

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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.

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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.

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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.

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