Question

Find the equation of the normal to y=log base e (x+2) which is parallel to the line with equation y+3x-5=0

Answer 1

@kropot72

Answer 2

this is how i did it

Answer 3

y+3x-5=0
y=-3x+5
gradient = -3

find the derivative of y=log base e (x+2):

y'=1/(x+2)

therefore derivative of y is equal to the gradient:

1/(x+2)=-3
x=-7/3

Answer 4

am i solving it correctly? @kropot72

Answer 5

the problem is i get an error for solving for y when i plug in the value of x=-7/3

Answer 6

do you think there is something wrong in the question?

Answer 7

help @ganeshie8

Answer 8

you set the tangent line equal to gradient, not the normal

Answer 9

ok

Answer 10

the slope of tangent is always positive on a log function

anyway, the normal slope is the opposite reciprocal to the tangent

--> -(x+2) = -3

Answer 11

so -1/m is the gradient of the normal

Answer 12

correct

Answer 13

ok now i get it

Answer 14

so m in this case is -1/3

Answer 15

The product of the slope of y = log base e (x + 2) and -3 must equal -1.
\[\Large \frac{1}{x+2} \times -3 = -1\]
Solve for x. Then plug the value of x into y = log base e (x +2). The resulting y value is then used to find the required equation , which is of the form y = -3x + b

Answer 16

will it become -1/3=1/(x+2)???

Answer 17

@Chad123
no m = -3 (slope of parallel line)

gradient of normal = -1/y'

Answer 18

\[\Large \frac{1}{x+2}=\frac{1}{3}\]

Answer 19

gradient of normal then is = -1/(1/(x+2))

Answer 20

am i correct?

Answer 21

Surely the gradient of the normal must equal -3.

Answer 22

can someone please show me how it is done step by step because im kind of lost

Answer 23

follow @kropot72 posts

Answer 24

The value of x when the normal is parallel to y + 3x - 5 = 0 is 1.
Put x = 1 into y = log base e (x + 2) to get y = ln 3 = 1.099.
Plug y = 1.099 and x = 1 into
y = -3x + b
and find the value of b. Then you will have the solution.

Answer 25

where did x=1 come from?

Answer 26

The solution to:
\[\Large \frac{1}{x+2}=\frac{1}{3}\]

Answer 27

1/3 came from plugging m=-3 into -1/m yielding 1/3

Answer 28

am i right?

Answer 29

Looks right to me, although I didn't put it that way.

Answer 30

solving for x gives 1

Answer 31

now it makes sense

Answer 32

thank you so much :)

Answer 33

Gr8! You're welcome :)

Answer 34

ill still have to solve the equation of the normal which i would probably know how to solve

Answer 35

Just do exactly as my previous post says:
"The value of x when the normal is parallel to y + 3x - 5 = 0 is 1. Put x = 1 into y = log base e (x + 2) to get y = ln 3 = 1.099. Plug y = 1.099 and x = 1 into y = -3x + b and find the value of b. Then you will have the solution."
Note: We already know that the form of the equation of the normal is:
y = -3x + b

Answer 36

So you just need to find the value of b by substitution of x = 1 and y = 1.099.

Answer 37

im using the equation of the normal formula

y-f(a)=(-1/f'(a)) (x-a)

Answer 38

You don't need to use that equation at this point. Please just do the substitution as I described.

Answer 39

ust plug x = 1 and y = 1.099 into y = -3x + b. Then you will have the value of b, and the equation of the normal

Answer 40

i will

Answer 41

thank u :)

Answer 42

you were very helpful

Answer 43

np