Tangent to $2^x$ passing through point (1.0) I've been asked to find a tangent to $2^x$, where the line goes through point (1,0). Know it's important to realise that it's not the line at the point (1.0) im trying to find. 
If we say that x=a, 
$2^a *ln(2)+2^a-ln(2)*2^a*a$
Which we can solve for a, however I do not understand how we can go from this to $2eln2(x-1)$ Which is the correct answer. 
'sorry for format,  on holiday and only have my phone'
 A: If we have $f(x)=2^x$, it follows that $f'(x)=\ln(2)2^x$.
Taking the general formula for a tangent, $(x-x_1)f'(x)=(y-y_1)$, simplifying & factorising, we have that 
$$(x-1)\ln(2)2^x=2^x\quad\Rightarrow\quad(x-1)\ln(2)2^x-2^x=0$$
$$\Rightarrow\quad 2^x\:\left(\ln(2)\,x-\ln(2)-1\right)=0$$
Which has the single solution $x=1+\frac{1}{\ln(2)}$
$f'(x)$ at this point is $\ln(2)2^{1+1/\ln(2)}=2e\ln(2)$
If we input this into our general tangent function, we get $(x-1)\cdot2e\ln(2)=y$
Which is what we wanted. Hope I helped.
A: Since $y'=2^x\ln 2$,  the equation of the tangent line at $(a,2^a)$ is
$$y-2^a=2^a(\ln 2)(x-a)\tag1$$
Since setting $x=1,y=0$ gives you
$$0-2^a=2^a(\ln 2)(1-a),$$
you can divide the both sides by $2^a\gt 0$ to get $$a=1+\frac{1}{\ln 2}.$$ Then, you can plug it in $(1)$ to get the answer.
A: Assuming the curve to be $y=2^x,$ the parametric point being $(a,2^a)$
$$\frac{dy}{dx}_{(\text{ at } x=a)}=2^a\ln 2$$
Now the gradient of the straight line connecting $(a,2^a);(1,0)$ is $\displaystyle\frac{2^a}{a-1}$
Now  for tangency, these two must be identical.
$$\implies2^a\ln a=\frac{2^a}{a-1}\iff(a-1)\ln 2=1$$
Hope you can take it home from here.
