Sister Clo
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Please explain how to sketch the graph of y=ln(2x-3). Thanks.
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AHappyStudent
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(Original post by Sister Clo)
Please explain how to sketch the graph of y=ln(2x-3). Thanks.
Start by looking at the equations

y=ln(x)
y=ln(2x-3)

What transformation/s turn ln(x) into ln(2x-3)

Then apply these to the graph

It's also a good idea to sub in y=0 or x=0 and solve for x or y respectively to find some intercepts to guide you
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RDKGames
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(Original post by Sister Clo)
Please explain how to sketch the graph of y=ln(2x-3). Thanks.
Consider what transformations take place when you go from \ln(x) (which I hope you know how to draw) to \ln(2x-3). Then consider how it would look like after these transformations, keeping in mind the critical points.
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Sister Clo
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(Original post by RDKGames)
Consider what transformations take place when you go from \ln(x) (which I hope you know how to draw) to \ln(2x-3). Then consider how it would look like after these transformations, keeping in mind the critical points.
Thanks. I transform by 2x (stretch in the x direction, so half the x-values), then translate 3 to the right. Doesnt seem to be the same when using a grapher.
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RDKGames
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(Original post by Sister Clo)
Thanks. I transform by 2x (stretch in the y direction), then translate 3 to the right. Doesnt seem to be the same when using a grapher.
Why the y-direction? You're applying these onto y=\ln(x), not y=x. If it were the latter than it would be true, however alternatively it is also a stretch by s.f. 1/2 in x-direction.

Before that happens though, I recommend you translate it by 3 to the right because \ln(x) \mapsto \ln(x-3) is a translation to the right by 3 units, and then \ln(x-3) \mapsto \ln(2x-3) is a stretch by s.f. 1/2 along the x-axis.
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Sister Clo
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(Original post by AHappyStudent)
Start by looking at the equations

y=ln(x)
y=ln(2x-3)

What transformation/s turn ln(x) into ln(2x-3)

Then apply these to the graph

It's also a good idea to sub in y=0 or x=0 and solve for x or y respectively to find some intercepts to guide you
Thanks - I stretch in the x-direction by scale factor half then translate 3 to the right. Seems to be different using graph plotter
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Sister Clo
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(Original post by RDKGames)
Why the y-direction? You're applying these onto y=\ln(x), not y=x. If it were the latter than it would be true, however alternatively it is also a stretch by s.f. 1/2 in x-direction.

Before that happens though, I recommend you translate it by 3 to the right because \ln(x) \mapsto \ln(x-3) is a translation to the right by 3 units, and then \ln(x-3) \mapsto \ln(2x-3) is a stretch by s.f. 1/2 along the x-axis.
Thanks yet again - I think the issue I have problem getting my head around is why translate first, when I would do the 2x first for all other graphs without a problem.
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RDKGames
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(Original post by Sister Clo)
Thanks yet again - I think the issue I have problem getting my head around is why translate first, when I would do the 2x first for all other graphs without a problem.
You can do stretch first, just depends on the preference. You apply transformations to x, so if you do stretch by s.f. 1/2 first you end up with \ln(2x) and now you need to translate by an appropriate a such that \ln(2(x-a)) is equal to \ln(2x-3) - clearly a\neq 3 here like it could be in the first scenario.
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Sister Clo
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(Original post by RDKGames)
You can do stretch first, just depends on the preference. You apply transformations to x, so if you do stretch by s.f. 1/2 first you end up with \ln(2x) and now you need to translate by an appropriate a such that \ln(2(x-a)) is equal to \ln(2x-3) - clearly a\neq 3 here like it could be in the first scenario.


Thank you. Much appreciated.
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