solve for $\int_{0}^{{\alpha}{b}}(a^x-1)dx=\int_{{\alpha}{b}}^{b}(a^x-1)dx$ I am sitting with a problem and my calculus is a bit (ok very) rusty.
$\int_{0}^{{\alpha}{b}}(a^x-1)dx=\int_{{\alpha}{b}}^{b}(a^x-1)dx\\
0<\alpha<1\\
b\geq1$
Solve for a.
any help would be greatly appreciated...
 A: You can write $a^x$ as $e^{x\ln a}$. Using this trick you can integrate $a^x$ as follows
$$
\int a^x\,dx = \int e^{x\ln a}\,dx = \frac{e^{x\ln a}}{\ln a} + C = \frac{a^x}{\ln a} + C
$$
Now apply the fundamental theorem of calculus to evaluate both integrals. You will end up with an equation involving $a$, however, I don't believe it can be solved analytically. Note that $a = 1$ is a solution since both integrals evaluate to zero, albeit, not a very interesting one.
A: (Edited with a solution for $\alpha <\frac 12$)

We have equation of the form $\int_0^{\alpha b}f(x)\ dx = \int_{\alpha b}^b f(x)\ dx$ and that is equivalent to $$2F(\alpha b) = F(b) + F(0)$$ where $F'=f$. Since we know that $(a^x)'=a^x\ln a$, it easily follows that $$\int (a^x-1)\ dx = \frac{a^x}{\ln a}-x + C$$ so we need to solve equation
$$2\left(\frac{a^{\alpha b}}{\ln a}-\alpha b\right) = \frac{a^b}{\ln a}-b + \frac 1{\ln a}$$ If we multiply it by $\ln a$ and substitute $t = b\ln a$, we get \begin{equation}
e^{\alpha t} -\alpha t =\frac{ e^t-t+1}2\tag{1}
\end{equation}
Let us define $g(x)=e^x-x$, so, $(1)$ becomes
\begin{equation}
g(\alpha t) = \frac{g(t)+1}2\tag{2}
\end{equation}
Notice that $g$ is a convex function and $g(x)\geq 1$ (take first derivative, equate it to $0$ and find minimum). Now, since $0<\alpha<1$, by convexity we have $$g(\alpha t) = g(\alpha t + (1-\alpha)\cdot 0)\leq \alpha g(t)+(1-\alpha)g(0)\tag{3}$$ Thus, if we assume $t$ solves $(2)$, and then substitute $(2)$ into $(3)$, after simplification we get 
\begin{equation}
(1-2\alpha)g(t) \leq 1-2\alpha\tag{4}
\end{equation}
Now, if $\alpha <\frac 12$, then $(4)$ implies that $g(t)\leq 1$, but $1$ being minimum of $g$ then implies $g(t) = 1$, i.e. $t=0$ and consequently $a=1$. Easily enough one check that $a=1$ indeed solves the original equation for any $\alpha$, and for $\alpha <\frac 12$ it is unique solution.

What about $\alpha \geq \frac 12$? Returning to $(1)$, we can solve it for $\alpha$. We have equation of the form $e^x-x=c$, so let us solve that one first:
$$
\begin{align}
e^x-x=c &\implies e^x = x+c\\
&\implies e^{-c}e^{x+c} = x+c\\
&\implies e^{-c} = (x+c)e^{-(x+c)}\\
&\implies -e^{-c} = -(x+c)e^{-(x+c)}\\
&\implies W(-e^{-c}) = -x-c\\
&\implies x = -W(-e^{-c})-c\\
\end{align}$$
Setting $x=\alpha t$ and $c=\frac{e^t-t+1}2$ we get $$\alpha = -\frac 1 t\left[ W\left(-\exp\left(-\frac{e^t-t+1}2\right)\right)+\frac{e^t-t+1}2 \right]$$ and when we remember our substitution $t=b\ln a$, finally it comes to $$\alpha = -\frac 1 {b\ln a}\left[ W\left(-\exp\left(-\frac{a^b-b\ln a+1}2\right)\right)+\frac{a^b-b\ln a+1}2 \right]$$
(Note: $W$ is Lambert W function.)
We can then plug in, for example, $t=2$ and get $\alpha\approx 0.779429$, so, for that $\alpha$ we have another solution $a=e^{\frac 2b}$. In conclusion, solution need not be unique for $\alpha \geq \frac 1 2$, but I don't know how to solve it.
