Question on Log Function (1 Viewer)

[Praer]

Find the equation of the tangent to y = logx at the point where e^2.
Can't seem to find the equation -.-"
The answer is supposed to be ( x - e^2 * y +e^2 = 0 )

 

[Shadowdude]

What log is it? Natural log? Base 10?

 

[Aesytic]

first find dy/dx of y=logx
dy/dx = 1/x

i'm assuming you mean at the point where x=e^2, so sub this into dy/dx to find the gradient of the tangent to y=logx where x=e^2
.'. gradient of tangent = 1/e^2
now when x=e^2,
y = log(e^2)
= 2loge
= 2
so the point on the curve where x=e^2 is (e^2, 2)

now using point gradient formula with the point and gradient that we have found, we can find the equation of the tangent:
y - 2 = 1/e^2 * (x-e^2)
= x/e^2 - 1
multiply through by e^2:
y*e^2 - 2e^2 = x - e^2
x - y*e^2 + e^2 = 0 which is the equation of the tangent

 

[Praer]

first find dy/dx of y=logx
dy/dx = 1/x

i'm assuming you mean at the point where x=e^2, so sub this into dy/dx to find the gradient of the tangent to y=logx where x=e^2
.'. gradient of tangent = 1/e^2
now when x=e^2,
y = log(e^2)
= 2loge
= 2
so the point on the curve where x=e^2 is (e^2, 2)

now using point gradient formula with the point and gradient that we have found, we can find the equation of the tangent:
y - 2 = 1/e^2 * (x-e^2)
= x/e^2 - 1
multiply through by e^2:
y*e^2 - 2e^2 = x - e^2
x - y*e^2 + e^2 = 0 which is the equation of the tangent

Legend, i made a mistake when finding the y value -.-"
Tnx