With $a$ being a real number, the graphs of the two functions $y=2^{x-a}-1$ and $y=a\log_2(x-a+1)$ intersect in $A\in Ox$ and $B$. When the area of $OAB$ is $\dfrac{7a}{2}$, find the coordinates of the midpoint of $AB$.
It's easy to see that $A(a;0)$ but i can't find the point $B$, i tried to find the $x$-coordinate of $B$ by solving exponential-logarithms equation but with no success. Thank you if you can give me any clues.
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Consider that you look for the zeros of function $$f(x)=(2^{x-a}-1)-a\frac{ \log (x-a+1)}{\log (2)}$$ Easier is to let $x=t+a$ and to consider $$g(t)=2^t-1-a\frac{ \log (t+1)}{\log (2)}$$ Since $g(0)=0$, we have the first root $x=a$.
We have $$g'(t)=2^t \log (2)-\frac{a}{(t+1) \log (2)}$$ which cancels for $$t_*=\frac{W\left(\frac{2 a}{\log (2)}\right)}{\log (2)}-1$$ where $W(.)$ is Lambert function.
Using a Taylor expansion we have $$g(t)=g(t_*)+\frac 12 g''(t_*)\, (t-t_*)^2+O((t-t_*)^3)$$ and then the approximation $$t\sim t_*+\sqrt{-2 \frac{g(t_*)}{g''(t_*)}}$$
I think that now, you need to plot $t$ as a function of $a$.
Set the equations Equal to each other: $$2^{x-a}-1=a\log_2(x-a+1)$$ Add your real number $a$ and you should be fine.
I think the intersections are roughly $a$ and $2a$, which is a very approximate value for small $a$