# Solving the formula $x e^x = e$

If you intend to confirm the suggestion that if $x e^x = e$ after that $x = 1$, one means to do it is to simply show that (1) $x = 1$ is without a doubt a remedy of $x e^x = e$ by connecting the value in, which (2) $x = 1$ is the *only * remedy, for which you can say that $x e^x$ is a raising function on $x \ge 0$, and also the formula plainly has no remedy for $x < 0$.

If that makes it seems like you need to recognize the solution ahead of time prior to you can address for it, well, that is not much from the fact. As a whole, if the appropriate - hand side is an approximate actual number $y$ as opposed to $e$, after that you can not address $x e^x = y$ for $x$ in regards to primary functions. One actually *specifies * the Lambert W function as specifically the remedy of this formula.

Another means is considering this as asking where do the charts of the features offered by $f(x) = e^{x-1}$ and also $g(x) = \frac{1}{x}$ intersect. This is due to the fact that $$ xe^{x} = e \Leftrightarrow e^{x-1} = \frac{1}{x} \quad \text{for $x \neq 0$}$$ Then the solution you have would certainly represent the only junction factor $(1, 1)$. So if you have no suggestion of where to start you can simply attract the equivalent charts which can offer you a suggestion of what is taking place.

Then the charts possibly would certainly recommend to you the adhering to logic. If $0 < x < 1$ after that $$ \frac{1}{x} > 1 \quad \text{but} \quad e^{x-1} < e^{0} = 1$$ so their charts can not fulfill in the interval $]0, 1[$. After that if $x > 1$ we have actually $$\frac{1}{x} < 1 \quad \text{but} \quad e^{x-1} > e^{0} = 1$$ so once more there is no junction factor in the interval $]1, \infty[$.

This dismiss all the feasible remedies in addition to $x = 1$ which you currently recognized and also which can be presumed by considering the charts.

Assuming that we are just seeking actual remedies,

$$xe^x = e.$$

Note that $e^x$ is constantly favorable therefore no remedy for this formula is feasible for $x<0$ or $x=0$. As x rises, both x and also $e^x$ are purely raising. Consequently, there is just one value for which $xe^x = e$. We currently recognize that remedy to be $x=1$.

With logarithms.

To start with $x>0$, Then :

$xe^{x}=e \Rightarrow \ln (xe^{x})=\ln e \Rightarrow \ln x + x = 1 $.

So :

If $x>1$ after that $ \ln x >0$, hence $\ln x +x > 1+0 =1$ (opposition)

If $x<1$ after that $ \ln x < 0$, hence $\ln x +x < 1+0 =1$ (opposition)

Thus $x=1$

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