Intersection between a natural logarithmic function and its inverse function

Question

I've been stuck on this math problem for a while and it involves solving the intersections between a natural log and its inverse function.

The natural log function is: $\ln(x+2)+3$

The inverse function is: $e^{x-3}-2$

I understand that they reflect on the $y=x$ line, but when I write them down as: $\ln(x+2)+3 = e^{x-3}-2$, I can't go any further than $\ln(x+2)+5 = e^{x-3}$.

So I was wondering if there is anyway I can solve this without putting it into the graphics calculator.

Any help would be greatly appreciated.

Thank you in advance.

Answer 1

We better subtract both sides with $\ln(x+2)+5$, $$F(x)=\ln(x+2)-e^{x-3}+5$$ this kind of function is called a transcendental function, and we often use Newton's method
to find it's root, If you want high accuracy, it can be really hard to solve by hand, almost impossible. But if you only need accuracy at, say 2 decimal places, Newton's method works fine.

Answer 2

Knowing that you want to find the positive zero of function $$f(x)=\log(x+2)-e^{x-3}+5$$ the solution should be close to $x=5$ since $e^2 \sim 7.4$ and $\log(7) \sim 2$.

Let $x=t+5$ and make a series expansion around $t=0$; limited to two terms, this gives $$f(t)=\log (t+7)-e^{t+2}+5=\left(5-e^2+\log (7)\right)+\left(\frac{1}{7}-e^2\right) t-\left(\frac{1}{98}+\frac{e^2}{2}\right) t^2+O\left(t^3\right)$$

Using a series reversion $$t=\frac{7 \left(\log (7)-e^2+5\right)}{7 e^2-1}+\frac{7 (1+49e^2) \left(\log (7)-e^2+5\right)^2}{(7 e^2-1)^3}+\cdots=-0.0630678$$

So, as an approximation, $x=4.93692$ while the "exact" solution (given by Newton method) is $x=4.93685$.

For sure, we could do much better with one more term; the same process would lead to $x=4.936851$ to be compared the exact $4.936847$