Finding equation of straight line that is tangent to $y = 2^x$. Problem: Find an equation of the straight line that is tangent to $y= 2^x$ and that passes through the point $(1,0)$.
Attempt: Let $(a, 2^a)$ be the point of tangency. Now we have that $y' = 2^x \ln(2)$, which evaluated at the tangency point becomes $y' = 2^a \ln(2)$.
So the general equation of the tangent line is $y = 2^a \ln(2) (x-a) + 2^a$. If this line is to go through $(1,0)$, then we must demand that $0 = 2^a \ln(2) (1-a) + 2^a$. I rewrote this as $-2^a = 2^a \ln(2) (1-a)$, which simplifies to $ -1 = \ln(2) (1-a)$. Now I'm having trouble finding $a$ from this. If I distribute I get  $ a \ln(2) = \ln(2) + 1$ or  \begin{align*} a = \frac{ \ln(2) + 1}{\ln(2)} \end{align*}
Can I simplify this any further? Because my textbook says the tangent line has equation \begin{align*} y = 2e  \ln(2) (x-1) \end{align*} How can I find this answer?

To show the answers are the same, we need to show that if $a=\frac{\ln 2+1}{\ln 2}$ then $2^a=2e$.
Note that $a=1+\frac{1}{\ln 2}$, so 
$$2^a=2\cdot 2^{1/\ln 2}=2\cdot (e^{\ln 2})^{1/\ln 2}=2e.$$
