Given that $\int_{0}^{100} (a^x-1) \:\mathrm{d}x = 30$, how can I calculate $a$? How can I separate a from other constants so that I can evaluate a? I'm stuck at the bottom line as shown below.
\begin{align}
30& =\int_{0}^{100} (a^x-1) \: \mathrm{d}x
\\&= \left [ (x+1)a^{x+1}-x\right ]_{0}^{100} 
\\&= 101a^{101} - 100 - a
\end{align}
 A: Hints:
$$\int a^x=\frac{a^x}{\log a} +C\implies$$
$$\int\limits_0^{100}(a^x-1)dx=\left.\left(\frac{a^x}{\log a}-x\right)\right|_0^{100}=\ldots$$
A: Hint :
Rewrite $a^x$ as $e^{\large x\ln a}$, then
\begin{align}
\int_0^{100}(a^x-1)\ dx&=30\\
\int_0^{100}\left(e^{\large x\ln a}-1\right)\ dx&=30\\
\left[\frac{e^{\large x\ln a}}{\ln a}-x\right]_{x=0}^{100}&=30\\
\left[\frac{a^x}{\ln a}-x\right]_{x=0}^{100}&=30.\\
\end{align}
A: Start with $$\int (a^x-1) \:\mathrm{d}x = \frac{a^x}{\log (a)}-x$$ Then $$\int_{0}^{100} (a^x-1) \:\mathrm{d}x = \frac{a^{100}-1}{\log (a)}-100$$ and you want this result to be equal to $30$. So $$\frac{a^{100}-1}{\log (a)}-130=0$$ which cannot be solved using elementary functions. The only analytical solution involves Lambert function.
But, from a practical point of view, we can think that $a$ is close to $1$; then, a Taylor expansion of the last expression write $$0=-30+5000 (a-1)+\frac{492500}{3} (a-1)^2+O\left((a-1)^3\right)$$ So, using the first order approximation, we obtain $$a=\frac{503}{500}=1.0600$$ Using the second order expansion, we obtain $$a=\frac{9700+3 \sqrt{4470}}{9850} =1.00513$$ while the exact solution is $a=1.00505$. 
In fact, we could easily polish the root using Newton method starting at $a_0=1.005$. The successive iterates will then be :$1.00508$,$1.00505$ which is the solution for six significant figures.
If you want the analytical expression, it is $$a=\sqrt[100]{-\frac{13}{10} W_{-1}\left(-\frac{10}{13 e^{10/13}}\right)} \simeq 1.005049060013037195011879$$
Added later to this answer
It is interesting to notice that, if we start Newton procedure with the initial value proposed by Henry, that is to say $$a_0=2\times \left(\frac{130}{100}\right)^{1/100} - 1$$ the first iterate is equal to $1.0050504855941330185$ and the second iterate is $1.0050490600821571111$
