Find the equation of the tangent line to the curve y = e^x when it is parallel to 2x - y = 7

Question

y = e^x 2x - y = 7

m = 2

y' = e^x

e² = 2

x = ln (2)

e^ln(2)= 2

(ln 2,2)

y - y₁ = m(x - x₁)

y - 2 = 2(x - ln 2)

y = 2x - ln 2 + 2

y = 2x - 1.39 + 2 is the equation of the tangent line

am i correct?

Answer 1

put the linear function into slope-intercept form --> y = 2x - 7

for a line to be parallel, it's slope must be equivalent to the slope of the line that it's parallel to

So, take the derivative of e^x --> dy/dx = e^x

Find when e^x = 2

x = ln2

Then, plug this value back into the original equation, so y = e^ln2

y = 2

the point on the original function that we're at is (ln2, 2)

Create a line using y = mx + b

solve for b

2 = 2(ln2) + b

b = 2 - 2ln(2)

y = 2x + 2 - 2ln(2)

So yes, you're correct.

Answer 2

The answer is correct.

Some of your derivations are not correct, but that might just be a typo.