Question about exponential $2^{(x+3)^{(x+2)}} = 4^{x+2}$

Question

I have a problem with this equation:

! I manage to find the first solution, which is x =-1 but it seems that there is another solution, which is x ≈ -1.19896. How do we do this, I can't seem to find any thread about solving exponential equation where the base is a variable itself.

Thank you!

Answer 1

The values $x+2=1$ and $x+3=2$ are compatible giving the apparent equality $2^2=4$ hence $x=-1$ is a solution.

Taking into account the convexity of the exponential function ($(x+3)^{x+2}=e^{(x+2)(\ln(x+3)}$), if a line passing through a point on the curve is not tangent to the curve then this line cuts the curve in another point. We search for another point distinct of $(-1,2)$.

We have clearly the equivalent equation $$(x+3)^{x+2}=2(x+2)$$ Make $$\begin{cases}f(x)= (x+3)^{x+2}\\g(x)=2x+4\end{cases}$$ One has $$f\space’(x)= (x+3)^{x+1}(x+2+(x+3)\ln (x+3))$$ and it is apparent both that $ f\space’(-2)=0$ (which certainly corresponds to a minimum because of the convexity) and that $y=2x+4$ is not tangent to the curve. It follows there is a second solution so one has to solve a transcendental equation. We calculate approximations as usual. $$\begin{cases}f(-1)=g(-1)=2\\f(-1.1)=1.9^{0.9}\approx 1.78187\text{ and }g(-1.1)=1.8\\f(-1.2)=1.8^{0.8}\approx 1.60036\text{ and }g(-1.2)=1.6\end{cases}$$ since $$f(-1.1)\lt g(-1.1)\text { and }\space f(-1.2)\gt g(-1.2)$$ there is a point solution $x_0$ such that $$-1.2\lt x_0\lt-1.1$$ this is an small interval, of length $0.1$, that can be narrowed more and more by successive approximations. We stop here saying that the solution $x_0\approx -1.19896$ given by the O.P. can be approximated this way.

Answer 2

As said in comments and answer, th problem is to find the zero of function $$f(x)=(x+3)^{(x+2)}-2(x+2)$$ Expanding as a series around $x=-1$ (the known solution), we have $$f(x)=(2 L-1) (x+1)+\left(L^2+L+1\right) (x+1)^2+$$ $$\frac{1}{12} \left(4 L^3+6 L^2+12 L+3\right) (x+1)^3+O\left((x+1)^4\right)$$ where $L=\log(2)$.

Solving the quadratic and selecting the closest root $$x=- \frac{\left(4 L^3+12 L^2+18 L+9\right) -2 \sqrt{3} \sqrt{-5 L^4-2 L^3-9 L^2+12 L+6}}{4 L^3+6 L^2+12 L+3 }$$ which is $\color{red}{-1.20203}$.

Adding the next term, which is $$\frac{1}{24} \left(2 L^4+4 L^3+12 L^2+6 L+5\right)(x+1)^4$$ there is only one real root for the cubic which has been solved analytically using the hyperbolic solution. Its numerical value is $\color{red}{-1.19872}$.

Remember that the "exact" solution is $\color{blue}{-1.19896}$