How to find the number of solutions to this equation $\ln(x+2)=e^x-2 $?

Question

how do I solve the following equation (preferably no graph) : $\ln(x+2)=e^x-2 $

Answer 1

This equation can't be solved with elementary algebraic methods. Numerical methods can be used instead. You'll get 2 solutions.

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EDIT: If you want to find the number of solutions, you can move everything over to the left side to get the function $$f(x) = \ln(x+2)-e^x+2 \ , \ x\in\mathbb{R.}$$ The derivative is $f'(x) = 1/(x+2)-e^x$. You can use the Intermediate Value Theorem to proceed.

Answer 2

Let $f(x)=e^x-2$ and $g(x)=\ln(x+2)$.

Easy to see that $f(g(x))=g(f(x))=x$ and $f$ is an increasing function.

Thus, $f(x)=g(x)\Leftrightarrow f(x)=x$, which gives $$e^x=2+x,$$ which is a bit of easier.

Answer 3

Let $ f(x)=ln(x+2) $ , also we can prove that $ f $ is an increasing function thus it is $ 1-1 $ function so it's invertible and it is $ f^{-1}(x)=e^x-2 $. So $$ f(x)=f^{-1}(x)$$ and because $ f $ it is increasing $$ f^{-1}(x) = x \Leftrightarrow \\ e^x-2=x \Leftrightarrow \\ e^x-x-2=0 $$ Let $ g(x)=e^x-x-2 $ and we find that $ g'(x)=e^x-1 $. $ g $ is decreasing at $ (-\infty,0] $ and is increasing at $ [0,+\infty) $ (because $ g'(x)<0$ at $ (-\infty,0] $ and $ g'(x)>0$ at $ [0,+\infty) $). $$g(-2)g(0)<0 \\ g(0)g(2) < 0$$ and because $ g $ is continuous by bolzano's theorem $ g $ has one real root at $ [-2,0] $ and another one at $ [0,2] $. Using Newton's-Raphson Method:
For $ x_0=-2 \\x_1=x_0-{g(x_0)\over g'(x_0)} = -2-{g(-2)\over g'(-2)} \approx -1.843 \\ x_2=x_1-{g(x_1)\over g'(x_1)} = -1.843-{g(-1.843)\over g'(-1.843)} \approx -1.842\\ \vdots \\ x=-1.841$
For $ x_0=1 \\x_1=x_0-{g(x_0)\over g'(x_0)} = 1-{g(1)\over g'(1)} \approx 1.163 \\ x_2=x_1-{g(x_1)\over g'(x_1)} = 1.163-{g(1.163)\over g'(1.163)} \approx 1.146\\ \vdots \\ x=1.146$
so our two solutions are $ x=-1.841 $ and $ x=1.146 $ (approx.)