Suppose y= kt is a tangent to the function f(t) = 2^t, evaluate k? My thinking is because ln 2(2^t) set this equal to kt and solve. I cant?

hmmmm. my thinking is y=kt is a line that passess through (0,0) and 2^t is an exponential function.

so i would think find a tangent line on 2^t that passes through (0,0) then we know the slope is k.

ok so lets see..

so the tangent line would be of the form:

y=kt

y = ln2(2^t)t but we need a line so we have to find a t that fixes (2^t).

this tangent line is on the curve y=2^x and we can find the point of intersection by setting them equal.

(2^t) = ln2 (2^t)t

1 = ln2 t

t = 1/ln2.

So for y=kt, (k=ln2 (2^t))

y = ln2 (2^(1/ln2))t

I think we can simplify this line equation.

a^x is equal to e^(ln(a^x)).

y = ln2 (e^(ln(2^(1/ln2))))t

bring down exponent

y = ln2 (e^((1/ln(2))ln(2)))t

1/ln2 times ln2 = 1

y = ln2*e*t

here's a plot:

http://www.wolframalpha.com/input/?i=y+%3D+ln2*e*t...

so k is e*ln2

k ≈ 1.884

sorry, I meant to say thinking is because the derivative of 2^t = ln 2(2^t), set it = to kt and solve. Is this correct and how to solve for k? I would think the answer will have an e to make the tangent line slope with the curve 2^t at all points t, i am wrong there? Any way someone please help me solve this.