Determine the equation of the line that is tangent to y=8^x at the point on the curve where x=1/2
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.
To find the equation of the line that is tangent to the curve at the point where x=1/2, we need to determine the slope of the tangent line and the coordinates of the point of tangency.
1. Find the derivative of the function y=8^x with respect to x. This will give us the slope of the tangent line at any point on the curve.
Let's differentiate y=8^x:
Using the chain rule, we have:
dy/dx = ln(8) * 8^x
Therefore, the derivative of y with respect to x is dy/dx = ln(8) * 8^x.
2. Plug in x=1/2 into the derivative to find the slope of the tangent line at that point.
dy/dx = ln(8) * 8^(1/2)
Simplifying this further, we get:
dy/dx = ln(8) * 2√2
Thus, the slope of the tangent line at the point where x=1/2 is ln(8) * 2√2.
3. Now, we need to find the y-coordinate of the point of tangency. Substitute x=1/2 into the original function y=8^x:
y = 8^(1/2)
Simplifying this, we have:
y = √8
Therefore, the y-coordinate of the point of tangency is √8.
4. Now that we have the slope and a point on the line, we can use the point-slope form of a linear equation, y - y1 = m(x - x1), to find the equation of the tangent line.
Plugging in the values we found:
y - √8 = ln(8) * 2√2 * (x - 1/2)
This equation can be simplified further if desired.