omvamen71

2022-09-03

How can I solve this nature log equation?

$ln(x+2)={e}^{(x-4)}$

Is there any way to solve this equation without graphing or using GDC ?

Thank you

$ln(x+2)={e}^{(x-4)}$

Is there any way to solve this equation without graphing or using GDC ?

Thank you

Quinn Alvarez

Beginner2022-09-04Added 13 answers

The presence of two layers of exponentials does not give any hope for analytical solutions and numerical methods are required.

Consider

$$f(x)={e}^{x-4}-\mathrm{log}(x+2)$$

$${f}^{\prime}(x)={e}^{x-4}-\frac{1}{x+2}$$

$${f}^{\u2033}(x)={e}^{x-4}+\frac{1}{(x+2{)}^{2}}$$

The first derivative cancels for

$${x}_{\ast}=W\left({e}^{6}\right)-2\approx 2.49666$$

($W(z)$ being Lambert function); at this point

$$f({x}_{\ast})=\frac{1}{W\left({e}^{6}\right)}+W\left({e}^{6}\right)-6\approx -1.28095<0$$

On the other side, by inspection, $f(-1)=\frac{1}{{e}^{5}}$ 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.

Consider

$$f(x)={e}^{x-4}-\mathrm{log}(x+2)$$

$${f}^{\prime}(x)={e}^{x-4}-\frac{1}{x+2}$$

$${f}^{\u2033}(x)={e}^{x-4}+\frac{1}{(x+2{)}^{2}}$$

The first derivative cancels for

$${x}_{\ast}=W\left({e}^{6}\right)-2\approx 2.49666$$

($W(z)$ being Lambert function); at this point

$$f({x}_{\ast})=\frac{1}{W\left({e}^{6}\right)}+W\left({e}^{6}\right)-6\approx -1.28095<0$$

On the other side, by inspection, $f(-1)=\frac{1}{{e}^{5}}$ 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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