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$57-58=$ Graphing Logarithmic Functions Match the logarithmic function with one of the graphs labeled I or II.$$f(x)=\ln (x-2)$$

Graph II

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 3

Logarithmic Functions

Campbell University

McMaster University

Harvey Mudd College

Baylor University

Lectures

03:42

Graphing Logarithmic Funct…

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In Exercises 51 - 58, w…

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Match the logarithmic func…

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$49-58$ Combining Logarith…

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Match the exponential func…

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In Exercises $57-62,$ use …

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So if you want to try to find which graph corresponds with the the natural law function, well, the first thing you could do is look at the you the basic natural law function, which is why is equal to natural Lagerback's. And remember that natural log just means we have log base e g of X. And so this is equal to why so rewriting this an exponential form. What we have e to the y is equal to X. So if we were to sketch this year, we were two sketches. So this is X and this is why Well, we have our ascent tote at zero. And so we know that at Y is equal to zero. X is equal to one, and when X is equal to X is equal to okay, why is equal to one? Let's say this is one and so so our graph looks something like this. I graph looks something like this. And so if if we have a a natural log of of two minus X of or rather X minus two X minus two means we're shifting this graph to the right, since it's inside the natural log function and subtracting two. That means we're shifting this graph to the right to So this this becomes this graph now looks like something like, let's say this is our by access. Let's say this is our y axis here and this is our X axis. So you have X and why? Well, now instead of X is equal to one. Now we have X is equal to three. So that this is one. This is too. This is three. This is this is for Well well, then we have at at X is equal to three. Now we have a at X is equal to three. Now we have a a point here or an exit Herceptin. And we know that our intercept is going to be as X approaches positive to as X approaches. Positive too. We have this this to this in black here. So we have this this natural log of to minus two. So we approach to but we never really We never really hit this value because then we have natural long of syrup. We know that's not possible or that would that would give us undefined. And so So we know that we have a nascent towed here at at X is equal to two. So let me label these points here. So we have one. One, 23 four. And then then my next point here will be approximately approximately two plus heat because because it's not exactly seven, because, remember, he is approximately 2.7. So if we're adding to we're adding to this would be approximately 4.7. So if this is five, this is five, then our 4.7 and be somewhere right here. And this would be X is equal to or rather why is equal to one. And so if this is, why is equal to one Well, our point would lie somewhere, right? Let's say here and so now our graph looks something like, Remember, we have this vertical Assen tote, and then our graph increases paragraph. Looks like this. So this is the second graph. This looks like our second graph here. So this is our solution. Graft to graph two corresponds to the natural log function that we were given to

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