omvamen71

2022-09-03

How can I solve this nature log equation?
$ln\left(x+2\right)={e}^{\left(x-4\right)}$
Is there any way to solve this equation without graphing or using GDC ?
Thank you

Quinn Alvarez

The presence of two layers of exponentials does not give any hope for analytical solutions and numerical methods are required.
Consider
$f\left(x\right)={e}^{x-4}-\mathrm{log}\left(x+2\right)$
${f}^{\prime }\left(x\right)={e}^{x-4}-\frac{1}{x+2}$
${f}^{″}\left(x\right)={e}^{x-4}+\frac{1}{\left(x+2{\right)}^{2}}$
The first derivative cancels for
${x}_{\ast }=W\left({e}^{6}\right)-2\approx 2.49666$
($W\left(z\right)$ being Lambert function); at this point
$f\left({x}_{\ast }\right)=\frac{1}{W\left({e}^{6}\right)}+W\left({e}^{6}\right)-6\approx -1.28095<0$
On the other side, by inspection, $f\left(-1\right)=\frac{1}{{e}^{5}}$ is a small positive number; the second derivative being always positive, there are two roots on each side of ${x}_{\ast }$.
Let us use Newton method.
For the first root, let us select ${x}_{0}=-1$; the successive iterates will then be
${x}_{1}=-0.993216345093696$
${x}_{2}=-0.993192968155096$
${x}_{3}=-0.993192967879949$
which is the solution for fifteen significant digits.
For the second root, let select ${x}_{0}=5$ (twice the value of ${x}_{\ast }$); the successive iterates will then be
${x}_{1}=4.70009929479913$
${x}_{2}=4.64012704349585$
${x}_{3}=4.63807063896736$
${x}_{4}=4.63806831105211$
${x}_{5}=4.63806831104913$
which is the solution for fifteen significant digits.

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