Help Basic logarithmic differentiation question.

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To find the tangent line to the curve y=8^x at x=1/2, the derivative is calculated as dy/dx=8^x(ln8), yielding a slope of approximately 5.9 at that point. The corresponding y-value is 8^(1/2)=2.8. The equation of the tangent line is derived using the point-slope form, resulting in 5.9x - y - 0.15 = 0. However, the correct answer involves simplifying expressions without rounding, leading to a more complex form involving roots and logarithms. It's essential to express all terms in their simplest form to avoid confusion in the final answer.
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Homework Statement


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

Homework Equations


Differentiation Rules
m=y2-y1/x2-x1

The Attempt at a Solution


y=8^x
dy/dx=8^x(ln8)
=8^0.5(ln8)
=5.9

y=8^x
=8^0.5
=2.8

5.9=y-2.8/x-0.5
5.9x-y-0.15=0 (My answer)

y=(6root2(ln2))x + root2(2-3ln2) (Real Answer)

How does this answer even make sense, I don't know where to go from here.
 
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Hints:
Don't round your answers; leave them the way they are.
ln8 = ln(2^3)=3ln2
root8 = 2root2
You're doing the problem correctly, but you may want to avoid rounding and reduce your expressions (for example, turn all the root8's into 2root2's).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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