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# C3 natural log graph help Watch

1. y = ln (2x)
y = (ln x)^2
y = ln (4-x)

I can't get my head around it. I subbed in values for previous questions and it gave me a zero which managed to work. But in these cases I don't know what to do. Can I get some help on these? Thanks
2. (Original post by Ze Witcher)
y = ln (2x)
y = (ln x)^2
y = ln (4-x)

I can't get my head around it. I subbed in values for previous questions and it gave me a zero which managed to work. But in these cases I don't know what to do. Can I get some help on these? Thanks
What is the question? Are you solving them? Sketching them? Finding intersections between them?
3. (Original post by ValerieKR)
What is the question? Are you solving them? Sketching them? Finding intersections between them?
Oh sorry. 'Sketch the following graphs stating any asymptotes and intersections with axes'.
4. (Original post by Ze Witcher)
Oh sorry. 'Sketch the following graphs stating any asymptotes and intersections with axes'.
When sketching ln(something) you just need to remember

ln(1) = 0
ln(0) tends to -infinity
ln(big numbers) tends to infinity, but the gradient is decreasing

for the squared one plot ln x and then plot it on that same graph - remembering that ln(e) = 1 and just making negative things positive, things between 0 and 1 closer to 0 and things above 1 bigger
5. (Original post by ValerieKR)
When sketching ln(something) you just need to remember

ln(1) = 0
ln(0) tends to -infinity
ln(big numbers) tends to infinity, but the gradient is decreasing

for the squared one plot ln x and then plot it on that same graph - remembering that ln(e) = 1 and just making negative things positive, things between 0 and 1 closer to 0 and things above 1 bigger
So are these general rules that I have to follow? And if it tends to -infinity, does it result in an asymptote?
6. (Original post by Ze Witcher)
So are these general rules that I have to follow? And if it tends to -infinity, does it result in an asymptote?
It's an easy technique to sketch any ln(f(x))

Yes
7. Ok, thank you

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