Trouble Isolating A Variable

Question

I ran across the following predicament in the course of solving a calculus problem. I'm embarrassed to admit I'm not sure how to proceed in order to solve for $x$.

$$e^x=6-x$$

I've tried taking the logarithm of each side yielding $x=\ln(6-x)$ where $x<6$. But this still leaves me in a position where I have trouble isolating $x$. Any help would be appreciated. Thanks!

Answer 1

The good news is it's not you-- it's the problem. You can't isolate the variable. The bad news is you can't isolate the variable.

But note $e^x$ is increasing. And $x-6$ is decreasing. So (in this case at least) the graphs cross and there will be exactly one $x$ where it is true.

So good news: You know there is one solution. Bad news: there is no algebraic way to express it.

But there are a lot of functions we can not express algebraically so we make up a function as a list of values to describe it. Example $\sin x$ is "the $y$ value of a point on a unit circle that has an interior angle of $x$". There's no way to algebraically solve that but we know it makes sense.

Likewise we know that for any $x$ there is a unique $y$ so that $ye^y = x$. We can't solve that algebraically but we know that $ye^y$ is increasing and has negative infinity and positive infinity as limits so we know $ye^y$ goes through every point exactly once so there is a unique $y$ so that $ye^y = x$.

So call this function the Lambert function $W(x)=y \iff ye^y = x$.

With this function in our arsenal:

We wont to get $\text{something about x}e^{\text{something about x}}= k$ so

$\text{something about x} = W(k)$. and then we can undo $\text{something about x}$.

$e^x = 6-x$

$\frac {e^x}{6-x} = 1$

$\frac 1{e^{-x}(6-x)} = 1$

$e^{-x}(6-x) = 1$

$e^{6-x-6}(6-x) = 1$

$e^{-6}*(6-x)e^{6-x}= 1$

$(6-x)e^{6-x} = e^{6}$

$6-x = W(e^{6})$

$x = 6-W(e^{-6})$

....

So $W(e^{6})$ is the value $y$ where $ye^y = e^{6}$ which we can only approximate. We can't solve algebraically.

Answer 2

You can write the equation in Lambert form: $xe^x = c$ with substitution and get an answer in terms of $W_0(c)$ where ${\displaystyle W_{0}(x)=\sum _{n=1}^{\infty }{\frac {(-n)^{n-1}}{n!}}x^{n}=x-x^{2}+{\tfrac {3}{2}}x^{3}-{\tfrac {8}{3}}x^{4}+{\tfrac {125}{24}}x^{5}-\cdots .}$

Here's the process of substitution and getting the answer:

$$e^x = 6-x$$

$$1= e^{-x}(6-x)$$

Now let $u = 6-x$ and $x = 6-u$.

$$1=e^{u-6}u$$

$$e^uu = e^6$$

$\therefore u = W_0(e^6)$

$x= 6-W_0(e^6)$

According to Wolfram Alpha, the sum does not converge and therefore, the solution does not exist.

Answer 3

For sure, the solution in terms of Lambert function is the only way to get an explicit formula.

If you do not use it, only numerical methods would work. Graphing the function, you see that the root is close to $\frac 32$. So, you could use Newton method and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.500000000 \\ 1 & 1.503340381 \\ 2 & 1.503335827 \end{array} \right)$$

Another way to do it is to Taylor expand $$e^x+x-6=\left(e^{3/2}-\frac{9}{2}\right)+\left(1+e^{3/2}\right) \left(x-\frac{3}{2}\right)+\frac{1}{2} e^{3/2} \left(x-\frac{3}{2}\right)^2+\frac{1}{6} e^{3/2} \left(x-\frac{3}{2}\right)^3+\frac{1}{24} e^{3/2} \left(x-\frac{3}{2}\right)^4+O\left(\left(x-\frac{3}{2}\right)^5\right)$$ and use series reversion to get $$x=\frac{3}{2}+\frac{t}{1+e^{3/2}}-\frac{e^{3/2} t^2}{2 \left(1+e^{3/2}\right)^3}+\frac{\left(2 e^3-e^{3/2}\right) t^3}{6 \left(1+e^{3/2}\right)^5}+\frac{\left(-e^{3/2}+8 e^3-6 e^{9/2}\right) t^4}{24 \left(1+e^{3/2}\right)^7}+O\left(t^5\right)$$ where $t=y-e^{3/2}+\frac{9}{2}$.

Now, make $y=0$ and get the same result (at least for ten significant figures).