Posted by Anonymous on .
Determine the equation of the line that is tangent to y=8^x at the point on the curve where x=1/2

calculus 
Reiny,
first the easy part,
when x = 1/2 y = 8^(1/2) = √8
so our point is (1/2, √8)
I will take ln of both sides
ln y = ln 8^x
lny = x(ln8)
(dy/dx) / y = ln8
dy/dx = ln8(8^x)
so at (1/2, √8)
dy/dx = (ln8)(√8)
so you have m = ln8(√8) and the point (1/2, √8)
find the equation of the line.