At the tangent point, the derivatives of the two functions are equal. It gives you an equation
= 2x. (1)
From this equation
x + a = . (2)
The next equation states that in the tangent point the values of "y" are the same for the two functions
ln(x+a) = x^2. (3)
Based on (2), replace (x+a) in equation (3) by . You will get then
= x^2, or
x^2 + ln(2x) = 0. (4)
Equation (4) is not algebraic and can not be solved by algebraic methods -- it only can be solved approximately
using numerical methods.
But the benefit of my solution is that the problem is reduced from two equations (1) and (2) to ONE SINGLE equation (4).
Now I go to website
https://www.wolframalpha.com/widgets/view.jsp?id=a7d8ae4569120b5bec12e7b6e9648b86
which provides free of charge online solver for non-linear scalar equations.
I input equation (4) into the specialized window and get the solution for x in a second
x = 0.419365.
Now from (2), a = = = 0.772914.
ANSWER. a = 0.772914.
Below is the plot to check the solution VISUALLY.
Plot y = ln(x+0.772914) (red) and y = x^2 (green)
