Y=ln(3-x) How to transform it? What is wrong with my logic?

[J_Ly08]

Hi, my name is Tim, I am new to this forum. Here's the description:

I don't understand how they've done it. I realize I can plot this, but what is the logic.

As a graph, I drew ln(-x) first, then added 3, so move left by 3 but that was not correct
I can see they've added 3 first, move left 3. Then they minus 1 times x. Or have they done it another way?
Why wasn't the first one correct, it makes sense to me

I also tried another approach which was ln(-(x-3)) so I first drew ln(x-3), and then I minus 1 times x, so I flipped it against the y axis. That also did not work, what is wrong with the logic on this one as well?

 

[dextercioby]

I really don't understand. The graph of \ln(3-x) is the curve which has exactly the same shape as \ln x, only that it is "crosses" the Ox axis in x=2 and diverges rapidly to -\infty in x=3.

In other terms, it the graph of \ln -x (which you get from the graph of \ln x by rotating it around the Oy axis) shifted rightwards on the Ox axis by 3 units.

 

[J_Ly08]

Thanks for the reply. Can you please explain/reword

In other terms, it the graph of ln−x (which you get from the graph of lnx by rotating it around the Oy axis) shifted rightwards on the Ox axis by 3 units.

I'm looking at it as transformation, so just like y=(x^2 + 3) moves x^2 up 3 units. Why isn't y=ln(3-x) moving ln(-x) 3 times to the left? and also to the last part of my question of my first post, if you write it as ln(-(x-3)) why wouldn't that work?

much appreciate for the help

 

[NascentOxygen]

I'm looking at it as transformation, so just like y=(x^2 + 3) moves x^2 up 3 units. Why isn't y=ln(3-x) moving ln(-x) 3 times to the left?

Why not? Because ln(3-x) is not the same as (ln(-x)) - 3

You should be able to show, viewing it as the sum of two logarithms,
ln(-x) - 3 = ln( (-x) / (e³) )

 

[Mark44]

Hi, my name is Tim, I am new to this forum. Here's the description:

I don't understand how they've done it. I realize I can plot this, but what is the logic.

As a graph, I drew ln(-x) first, then added 3, so move left by 3 but that was not correct

Here's why.

Basic function: y = ln(x)
Reflection across y-axis: y = ln(-x)
Translation: y = ln(-x + 3) = ln(-(x - 3))
Notice that this is a translation to the right by 3 units of the graph of y = ln(-x).

I can see they've added 3 first, move left 3. Then they minus 1 times x. Or have they done it another way?
Why wasn't the first one correct, it makes sense to me

I also tried another approach which was ln(-(x-3)) so I first drew ln(x-3), and then I minus 1 times x, so I flipped it against the y axis. That also did not work, what is wrong with the logic on this one as well?

Draw the basic untransformed graph, y = ln(x)
Reflect across the y-axis to get y = ln(-x)
Translate right by 3 units to get y = ln(-(x - 3)

On the first graph, one point is (1, 0).
On the reflected graph, this point moves to (-1, 0) [Check: ln(-(-1)) = 0]
On the translated graph, this point (in previous step) moves to (2, 0) [Check: ln(-(2 - 3)) = ln(-(-1)) = ln(1) = 0.

 

[J_Ly08]

Thanks for the responses.

After reading mark's reply its starting to clear up more. Much appreciate the help :)