**Outline :**

- Review -- Solving problems using
*ln* - Graphing The Exponential Functions
- Properties of The Exponential and Natural Log Functions
- Graphing the Natural Log Functions
- Introduction to Exponential Modeling

**Review -- Solving**

**problems**

**using**

*ln*:*ln5(3^x) = ln4^(x-1)*

*ln5 + xln3 = (x-1)ln4*

*2.*Distribute or multiply to get rid of the brackets.

*ln5 + xln3 = xln4 - ln4*

*3.*Move '

*x*' to one side then factor.

*ln5 + ln4 = xln4 - xln3*

*ln5 + ln4 = x(ln4 - ln3)*

You can go straight from step 2 to factoring like the image to the left, I'm just showing every step, because I don't want anyone to get lost (I get lost so easily so..)

4. As uhm, Justus would say (at least, I hear it from him a lot), next you algebra it.

*(ln5 + ln4) / (ln4 - ln3) = x*

Don't forget to give them 'homes' when you're writing it out. You lose marks if you don't, AND PLUS, it makes things look confusing, so you start making little mistakes.

5. Finally, if you want to look smart (8D) you can apply the the product and quotient laws to 'clean it up' a little. (I just figured this out now, I didn't really get why all of a sudden the numbers changed while we were doing it in class. AND THAT'S WHY WE REVIEW OUR NOTES GUYS) Therefore, you're final answer would look something like this!

*x = ln20 / ln(4/3)*

You don't have to do this btw, especially if you're not absolutely sure what's going on. If you try it on the final exam and get it wrong, you lose a

**full**mark, because it's a concept error. But you can go ahead and practice during class and on tests. =)

The other problem we looked at, we did last class also, so I won't go over that. You can just scroll down a little and check out Joseph's last post!

**Graphing The**

**Exponential Functions :**

As you all know, *e *is The exponential function. Above is what *y = e^x - 2, y = -e^-x, *and *y = I e^x - 1 I *looks like, with asymptotes *y = -2, y = 0 *and *y = 1* respectively. Remember to always indicate where the asymptotes are on the graphs. You can't just assume that they'll know where it is, even though it's really obvious.

The negative symbol [ *y = -e^-x *] in front of the *e* flips the graph horizontally, while the negative symbol in front of the exponent *x* flips the graph vertically. The absolute value symbol [ *y = I e^x - 1 I* ] means no value below -1 exists, thus flipping that side of the graph over the x-axis, changing its asymptote from -1 to 1.**Properties of The Exponential**

**and Natural Log Functions :**

Just in case you can't make that out..

y = e^x

y = e^x

**domain --**all permissible x-values of a function on a graph

*( -[infinity symbol], [infinity symbol] )*

**range --**all permissible y-values of a function on a graph

*( 0, [insert infinity symbol here] )*

**roots --**the point where a graph intersects the x-axis

*none*

**y-intercept --**the point where a graph intersects the y-axis

*y = 1*

**increasing or decreasing --**function with a graph that goes up as it's followed from left to right ; function with a graph that goes down as it's followed from left to right

*increasing*

**concavity --**concave up: a graph with an opening that's facing up ; concave down: a graph with an opening that's facing down

*concave up*

**asymptotes --**a line in which the function on a graph approaches infinitesimally closely but never meets

*y = 0*

y = lnx

y = lnx

**domain**

*--*all permissible x-values of a function on a graph

*( 0, [insert infinity symbol here] )*

**range**

*--*all permissible y-values of a function on a graph

*( -[infinity symbol], [infinity symbol] )*

**roots --**the point where a graph intersects the x-axis

*x = 1*

**y-intercept**

*--*the point where a graph intersects the y-axis

*none*

**increasing or decreasing**

*--*function with a graph that goes up as it's followed from left to right ; function with a graph that goes down as it's followed from left to right

*increasing*

**concavity**

*--*concave up: a graph with an opening that's facing up ; concave down: a graph with an opening that's facing down

*concave down*

**asymptotes**

*--*a line in which the function on a graph approaches infinitesimally closely but never meets

*x = 0*

**Graphing the Natural**

**Log Functions :**

*lnx*is the natural log function. Above are the graphs of

*y = ln(-x)*and

*y = -lnx + 2.*Recalling our knowledge of transformations, we were able to determine that [

*y = ln(-x)*] the negative symbol in front of the

*x*flips the graph along the y-axis. The negative symbol [

*y = -lnx + 2*] in front of the

*ln*flips the graph vertically.

**Introduction to**

**Exponential Modeling :**

So yes, that's the end, of that. The next scribe will be Jamie, I think she's the last scribe of this cycle? Yes? No? Well anyways, uhh, yeah. That's it. Test on Thursday-ish next week guys. Don't forget to BOB and to update your delicious .. things.

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