Maths C3 - Drawing Inverse Exponential Function graphs.. Help?

So yet again I'm struggling to self-teach Maths...

I find the Edexcel C3 Modular Maths Textbook difficult to learn from which ends up leaving gaps in my knowledge of the subject.

How would you go about drawing Inverse Exponential Functions onto a graph?
The example I'm struggling with atm is...

y=\ln (3-x)

I know that this is the Inverse of y=e^(-x+3) but I am still having trouble drawing it without having to draw the non-inverse exponential function first!! :frown:

(edited 7 years ago)

Reply 1

Kevin De Bruyne

Original post by Philip-flop

y=\ln (3-x)

I know that this is the Inverse of

y=e^(-x+3)

but I am still having trouble drawing it without having to draw the non-inverse exponential function first!! :frown:

Seems easier just to draw ln(3-x) instead of finding the inverse and reflecting it, no?

Reply 2

NotNotBatman

Original post by Philip-flop

y=\ln (3-x)

I know that this is the Inverse of

y=e^(-x+3)

but I am still having trouble drawing it without having to draw the non-inverse exponential function first!! :frown:

Consider the graph of

ln(x)

and apply transformations.

Reply 3

ValerieKR

log(0.000000001) = a very negative number
log(1) = 0
as a tends to infinity log(a) tends to infinity and the derivative of log(a) tends to 0

you can sketch log(x) using those four facts, then apply a couple of transformations

(edited 7 years ago)

Reply 4

Philip-flop

Original post by SeanFM

Seems easier just to draw ln(3-x) instead of finding the inverse and reflecting it, no?

Yeah that's what I want to be able to do. But the book doesn't really explain how to achieve this :frown:

Original post by NotNotBatman

Consider the graph of

ln(x)

and apply transformations.

Yeah but I don't really understand how to draw Natural Log graphs in the first place :frown:

Reply 5

Kevin De Bruyne

Original post by Philip-flop

Yeah that's what I want to be able to do. But the book doesn't really explain how to achieve this :frown:

Yeah but I don't really understand how to draw Natural Log graphs in the first place :frown:

As transformations go I think at A-level ln(3-x) is probably one of the most difficult ones I think.

You just have to start with lnx (you should become familiar with it) and then apply transformations.

Also in general with any graphs, if you know/can guess the shape then all you need are some points and you're set.

Eg thinking of ln(3-x) as a function.. what's the domain? You also know that ln(...) = 0 at some point, so what point is that at? What happens when x is very large (positive/negative? only one of those is in the range though) you can ask these kind of questions for any graph.

(edited 7 years ago)

Reply 6

Philip-flop

Would I be right in thinking that if... f(x) = lnx then f(-x+3) = ln(-x+3) ?? <<< also written as ln(3-x)
So I could just make my transformations that way?

Reply 7

NotNotBatman

Original post by Philip-flop

Yeah that's what I want to be able to do. But the book doesn't really explain how to achieve this :frown:

Yeah but I don't really understand how to draw Natural Log graphs in the first place :frown:

Do you know the graph of

e^x

? As

ln(x)

is it's inverse, it's a reflection in the line y=x. It should be in the textbook, or you can use an online grapher like desmos.

Reply 8

Philip-flop

Original post by NotNotBatman

Do you know the graph of

e^x

? As

ln(x)

is it's inverse, it's a reflection in the line y=x. It should be in the textbook, or you can use an online grapher like desmos.

Yes I know that e^x and that ln(x) is it's inverse. But for some reason I'm still having trouble :frown:

I have been using Desmos to try and make more sense of it :smile:

Reply 9

B_9710

Original post by Philip-flop

Would I be right in thinking that if... f(x) = lnx then f(-x+3) = ln(-x+3) ?? <<< also written as ln(3-x)
So I could just make my transformations that way?

Yes.

Reply 10

The_Big_E

Original post by Philip-flop

Would I be right in thinking that if... f(x) = lnx then f(-x+3) = ln(-x+3) ?? <<< also written as ln(3-x)
So I could just make my transformations that way?

Exactly.

Reply 11

B_9710

To get from

\ln x

\ln (x+3)

we first translate the curve by vector

\begin{pmatrix} -3 \\ 0 \end{pmatrix}

. Can you work out what we do next to get to

\ln (3-x)

Reply 12

Philip-flop

Wonderful explanations everyone!! I think I'm finally starting to understand it :smile:

Reply 13

Philip-flop

For anyone who is interested. Here is my answer...
Ch.3 (C3) Exa6 Answer.jpg

Reply 14

Philip-flop

The only thing I have trouble with now is working out where Natural Log graphs cross the x-axis say for example...

\ln(2x)+3

Reply 15

ValerieKR

Original post by Philip-flop

The only thing I have trouble with now is working out where Natural Log graphs cross the x-axis say for example...

\ln(2x)+3

They cross the x axis when y=0
in this case
ln(2x) +3 = 0
ln(2x)=-3
2x=e^-3
x=(e^-3)/2

Reply 16

Philip-flop

Original post by ValerieKR

They cross the x axis when y=0
in this case
ln(2x) +3 = 0
ln(2x)=-3
2x=e^-3
x=(e^-3)/2

Thank you so much!! I had actually forgotten how to solve Logs for a second there! You helped a lot!! :smile:

Reply 17

simon0

Original post by Philip-flop

y=\ln (3-x)

I know that this is the Inverse of y=e^(-x+3) but I am still having trouble drawing it without having to draw the non-inverse exponential function first!! :frown:

As a side question, you might want to double check your inverse function.

Your original function (

ln(3-x)

) crosses the x-axis at

(2, 0)

.

Your inverse should then cross the y-axis at

(0, 2)

, however, your inverse crosses the y-axis at

(0, e^{3})

(edited 7 years ago)

Reply 18

Philip-flop

Original post by simon0

As a side question, you might want to double check your inverse function.

Your original function (

ln(3-x)

) crosses the x-axis at

(2, 0)

.

Your inverse should then cross the y-axis at

(0, 2)

, however, your inverse crosses the y-axis at

(0, e^{3})

No, in this case ln(3-x) is the inverse of an e^x graph. I wasn't trying to find the reverse of ln(3-x) I was trying to work out what ln(3-x) looks like on a graph.