How can I solve this nature log equation?

Question

$ln(x+2)=e^{(x-4)}$ Is there any way to solve this equation without graphing or using GDC ?

Thank you

Answer 1

As Ross Millikan answered, the presence of two layers of exponentials does not give any hope for analytical solutions and numerical methods are required.

Consider $$f(x)=e^{x-4}-\log (x+2)$$ $$f'(x)=e^{x-4}-\frac{1}{x+2}$$ $$f''(x)=e^{x-4}+\frac{1}{(x+2)^2}$$ The first derivative cancels for $$x_*=W\left(e^6\right)-2\approx 2.49666 $$ ($W(z)$ being Lambert function); at this point $$f(x_*)=\frac{1}{W\left(e^6\right)}+W\left(e^6\right)-6\approx -1.28095 < 0$$ 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 $x_*$.

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 $x_*$); 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.

Answer 2

I would be surprised. It becomes $x+2=e^{e^{x-4}}$ and the mix of polynomials and exponentials is usually intractable without the W function. Having two layers of exponential is worse. The fact that Alpha gives only numeric answers says it could not solve it algebraically, either. It finds $4.63806831104913$ and $-0.993192967879949$ as approximate solutions.