Solve: $\ln x = e^x-e^1$ for $x$

Question

I realise this might seem basic but the fact that this equation has both "ln" and e is confusing me because I cannot seem to make the equation have the same base:

$$\ln x = e^x-e^1$$

Any help would be greatly appreciated thank you!!

Answer 1

HINT: there are two positive real solutions which can be obtained by a numerical method,e.g. the Newton-Raphson method

Answer 2

As you can see ..One solution is obvious ..The other has to be found numerically...

Answer 3

Start at $x_1=1/2$, then do $x_{n+1}=\exp(e^{x_n}-e)$

Answer 4

You have been told already that numerical methods should be used for the non trivial root.

But you must be careful. Consider the function $$f(x)=\ln x - e^x+e$$ Its derivative $$f'(x)=\frac 1x-e^x$$ cancels at $x_*=W(1)$ where appears Lambert function. Numerically, this means $x_*\approx 0.567143$ which corresponds to a maximum by the second derivative test. So, in order to avoid problems (even very serious), you must search for a solution between $x=0$ and $x=x_*$.

Being lazy, let us start using $x_0=0.25$. Newton method will then provide the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.25000000000000000000 \\ 1 & 0.23234075865580407607 \\ 2 & 0.23326694867597147957 \\ 3 & 0.23326974797292171657 \\ 4 & 0.23326974799836732850 \end{array} \right)$$