An **exponential function** is \(f(x) = b^x\) where \(b\) is any positive constant.

- If \(b > 1\), then \(b^x\) is exponential growth.
- If \(b < 1\), then \(b^x\) is exponential decay.

There is a special constant \(e\) which is a natural base for exponential functions. The **natural exponential function** is \(e^x\), and it has the very important property that its derivative is itself: \[\frac{d}{dx} e^x = e^x.\] In class, we used this property to:

- Find the slope and the formula for the tangent line to \(y = e^x\) at \(x = 0\).

We got the tangent line \(y = x + 1\), which means that \(e^x \approx 1+x\) when \(x\) is close to zero. This led to the following formula, which is the definition of \(e\): \[e = \lim_{n \rightarrow \infty} \left( 1+ \frac{1}{n} \right)^n.\] We used a calculator to approximate this limit with n = 10,000,000 and got \(e \approx 2.71828\).

We solved the following problems:

Find \(\dfrac{d}{dx} x^2 e^x\).

Find \(\dfrac{d}{dx} e^{3x}\).

A boiling cup of water is set out to cool and has temperature \(T(t) = 20 + 80 e^{-0.3t }\) where \(t\) is time in minutes. What is the instantaneous rate of change in the cup’s temperature at time \(t = 0\).

The normal distribution in statistics is given by \(y = e^{-\tfrac{1}{2}x^2}\). Find \(y'\) and \(y''\), and use \(y''\) to find the inflection points.

I tried to finish by explaining the connection between the natural exponential function and compound interest, but I ran out of time. Here’s the idea: if you invest \(P\) dollars (\(P\) for principal) at an interest rate of \(r\) per year, then next year you will have a balance \(B\) given by \[B = P + r P = P(1+r)\] If you wait more than 1 year, your interest will compound, which means you will earn interest on the interest you’ve already earned. So your balance after \(t\) years will be \[B(t) = P(1+r)^t.\] Typically banks will pay out interest quarterly or monthly, so the actual formula will be: \[B(t) = P\left(1 + \frac{r}{n} \right)^{nt}\] where \(n\) is the number of times interest is compounded per year. In the limit as \(n\) goes to infinity, this formula simplifies to the formula for **continuously compounded interest**:

\[B(t) = P e^{rt}.\] This is because \(\left(1+\frac{r}{n} \right) \approx e^{r/n}\) when \(n\) gets big, so if you raise both sides to the \(nt\) power, you get the result.

A *logarithmic scale* is a number line with distances that represent multiplying by fixed factors. For example, here is a log scale where each space represents a factor of 2.

A logarithm is a function that does two things.

\(\log_b(x)\) tells you how many steps on a base b log-scale the number \(x\) goes away from 1.

\(\log_b(x)\) tells you the power you need to raise the base to in order to get \(x\).

We calculated the following logarithms right away, without a calculator:

\(\log_5(25)\)

\(\log_{10}(10,000)\)

\(\log_2(\tfrac{1}{2})\)

\(\log_3(\sqrt{3})\)

All logarithms (no matter what base) satisfy three **Logarithm Rules**

\(\log(ab) = \log(a) + \log(b)\)

\(\log(\tfrac{a}{b}) = \log(a) - \log(b)\)

\(\log(a^p) = p \log(a)\)

We used these rules to simplify the following expressions:

_2(36)-_2(9)

_5(500) - _5(4)

_b(b^x)$

The most important logarithm is the base-e logarithm, which is called the **natural logarithm**. It is written \(\ln(x)\) instead of \(\log_e(x)\). It is the inverse of the function \(e^x\), which means that \[\ln(e^x) = x ~~~~~ \text{ and } ~~~~~ e^{\ln x} = x.\]

You should also know what the graph of \(\ln x\) looks like.

We did the following natural logarithm problems in class:

Simplify \(\displaystyle \ln \left( \frac{x^2}{\sqrt{y}} \right)\).

Simplify \(\ln(\sqrt{e})\)

Simplify \(\displaystyle \ln \left( \frac{e^2}{\sqrt{e}} \right)\).

Solve \(e^x = 2\).

If I invest in a bond that grows 2% per year, how long will it take for the bond to double? To solve this problem we needed to solve \((1.02)^t = 2\). The trick is to take the natural log of both sides.

The population of the United States is currently 330 million and is growing at 0.5% per year. If this trend continues, how long until the population reaches 400 million?

Today we talked about exponential growth. We started with these problems.

If you invest money at a 5% annual interest rate, how many times bigger will your investment be after \(t\) years?

After 40 years, how many times larger is an investment that grew by 10% every year compared with one that only grew 5% every year (assuming the two investments started with the same initial investment).

Surprisingly 5% growth is not just half of 10% growth. Over time, 10% growth will end up being much more than double 5% growth. Exponential growth can be surprising unintuitive. We watched this video in class: The Most IMPORTANT Video You’ll Ever See which explains a good way to understand exponential growth.

We finished today by looking at why the **Rule of 70** from the video is true. It has to do with the linear approximation for the function \(\ln(1+x)\).

We covered several topics today.

\[\log_b(x) = \frac{\ln x}{\ln b}.\]

- Use a calculator to find \(\log_2(5)\).

\[ \dfrac{d}{dx} \ln x = \dfrac{1}{x}.\]

We used this to solve the following problems in class:

Find the derivative of \(y = x \ln x\).

Find the intervals of increase and decrease for \(y = x \ln x\).

Use the chain rule to find \(\dfrac{d}{dx} \ln (x^2+4)\).

Find \(\displaystyle \frac{d}{dx} \ln \left( \frac{\sqrt{x-5}}{x^3}{(x+1)^4} \right)\)$. This problem would be extremely hard with the chain rule, but if you use the rules for logarithms to simplify the function first, it isn’t that bad.

Find \(\displaystyle \frac{d}{dx} \ln \left( \frac{\sqrt{x}}{x+1} \right)\).

The last thing we talked about today was derivatives of exponential functions with other bases.

\[\dfrac{d}{dx} a^x = a^x \ln a.\]

If the population of a town is growing at 5% per year and is currently 3000, use the derivative to find the rate of change of the population 10 years from now.

If the price of an item is currently $3 and is decreasing by 5% per year, find a formula for the price as a function of time (in years) and then find the derivative to estimate how much the price will be decreasing 5 years from now.