Question

Find the equation of the normal at x=2 to the curve with equation y=In(2x-3)

I think I know how to differentiate it, but further than that, I'm stuck.

Answer 1

@campbell_st can you see if you can help me, or anyone else?

Answer 2

for logs and differentiating its f'(x)/f(x)

so \[dy/dx = 2/2x - 3\]

substitute x = 2 to find the gradient then m = 2/(2 x 2 - 3)

m = 2

substitute x = 2 into the original function to find the y value in the point

y = ln(2 x 2 - 3) y = 0 since ln(1) = 0

now you have the point of contact of the tangent (2, 0) and the tangent has a gradient of 2.

use the point slope equation

\[y - y _{1} = m(x - _{1})\]

gives

y - 0 = 2( x - 2)

then y = 2x - 4

Answer 3

point gradient formula

\[y - y _{1} = m(x - x _{1})\]

Answer 4

2y=2-x is the answer... I thought it was y=2x-4 aswell.?

Answer 5

lol... oops I thought it was the tangent.... but you need to find the normal... I should read the question more carefully.

ok... for a normal the gradient is -1/m where m is the gradient of the tangent.

so the normal has a gradient m= -1/2 but still passes through the point (2, 0)

put them into the point gradient formula and you will get

y = -x/2 + 1 multiply each term by 2 to get rid of the fraction

will give 2y = 2 - x

sorry about not reading the question correctly

Answer 6

thanks