I understand this is probably a duplicate question, but the answers on said duplicates don't make sense to me or don't seem to work for me.
I'm being asked to find the point of tangency and the equation for the tangent line.
So far I have that:
- $y'=ln(2)2^x$
- There is a point $P=(a, 2^a)$ which is where the aforementioned tangent line passes through the graph
- It passes through (1,0)
Using this information I tried to create the equation for the tangent line.
$0-2^a=ln(2)2^a(1-a)$
And then I end up with
$0=2^a(ln(2)(1-a)+1)$
And this is where I get lost, I can't seem to extrapolate any information about a point of tangency from this.
Best Answer
For real finite $a,2^a>0$
$$\implies(a-1)\ln2=1$$
$$a=?$$