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Home , Function (mathematics), Logarithm, Natural logarithm

Lecture 5 - , of a ,

5.1 Logarithms

Note the graph of ex

This graph passes the , so f(x) = ex is one-to-one and therefore has an . This is also true of f(x) = ax for any a > 0, a 6= 1.

More generally, for any a > 1 the graph of ax and its inverse look like this. If f(x) = ax, then we define the inverse function f −1 to be the with a, and write

−1 f (x) = loga(x) x Note that, since the of a is only the positive , the domain of loga(x) is all positive real numbers. The key property is:

b loga x = b ⇐⇒ a = x Examples: ? log10 10 = 1 10 = 10 ? log5 25 = 2 5 = 25 1 1 ? 1 log4 2 = − 2 4 = 2 1 ? 1 log5 125 = −3 5 = 125 ↑ ↑ log corresponding exponential equation

Log Rules

1. Most important: by the properties of inverse functions we have

x logb x logb(b ) = x and b = x The most important case of logs is when b = e. Log base e has a special name, in fact

we define loge x = ln(x). So the above becomes

ln(ex) = x and eln(x) = x

LEARN THIS!! The function ln(x) is known as the function, and ln(x) should be read as “the natural logarithm of x”. In , you may also hear me read this as “lawn x”, but this isn’t as standard.

Other rules: (I will state for ln, but they work for every log). Suppose that x, y > 0

2. ln(xy) = ln(x) + ln(y)

 x  3. ln y = ln x − ln y 4. ln(xy) = y ln(x)

Calculations: e3 ln(x) = eln(x3) = x3 1 ln = ln(e−1) = −1 e Rewrite the following: xy  ln = ln(xy) − ln(z) = ln(x) + ln(y) − ln(z) z Revisited: (p324 #77) A deposit of $1000 is made in an account that earns interest at a rate of 5% per year. How long will it take for the balance to double if interest is compounded annually? solution: From our earlier , our balance after t years is

 .05t A(t) = 1000 1 + = 1000(1.05)t 1 We are trying to find t such that A(t) = 2000. So we the formula equal to 2000:

2000 = 1000(1.05)t

2 = 1.05t We have to solve this for t. The general principle we use is that if we are trying to solve for a in the exponent, take log of both sides. So we get

ln(2) = ln(1.05t)

ln(2) = t ln(1.05) In this last we see the point of using logs - the exponent can be brought down and solved for. So, ln(2) t = ≈ 14.21 years. ln(1.05) ln(2) Note that this is a very typical test/exam question. The answer I would expect is t = ln(1.05) . Example 2: Same question, but now interest is compounded 10 times a year. solution: By our formula,

 .0510t A(t) = 1000 1 + = 1000(1.005)10t 10 Again, we solve A(t) = 2000. 2000 = 1000(1.005)10t 2 = 1.00510t ln(2) = ln(1.00510t) ln(2) = 10t ln(1.005) Therefore, ln(2) t = ≈ 13.9 years. 10 ln(1.005) Example 3: Same question, but now interest is compounded continuously. solution: By our formula, A(t) = 1000e.05t Again, we solve A(t) = 2000. 2000 = 1000e.05t 2 = e.05t ln(2) = ln(e.05t) ln(2) = .05t Therefore, ln(2) t = ≈ 13.86 years. .05 Notice that continuous compounding gives a kind of “best case scenario” – no amount of compounding will get your money to double faster than approximately 13.86 years.

5.2 The and the Slope of Lines

When given a , we want to be able to calculate the slope of tangent lines.

The slope of the tangent line at a point will tell us how rapidly the function is increasing or decreasing. We use a to find the of tangent lines. A for a function is one that intersects it at at least two points. The idea is to approximate the tangent line using secant lines. The slope of the tangent line will be the limit of the slopes of the secants, if it exists. The slope of the secant line connecting x1 to x is

f(x1) − f(x)

x1 − x If we pick a point near to x and denote it x + ∆x, the slope of the secant is f(x + ∆x) − f(x) f(x + ∆x) − f(x) = (x + ∆x) − x ∆x We should think of ∆x as being small, and note it can be positive or negative. So ideally the slope of the tangent line is f(x + ∆x) − f(x) lim ∆x→0 ∆x So we make the following definition.

Definition 5.1 Let f(x) be a function. Then we define the derivative of f at x, denoted f 0(x), as f(x + ∆x) − f(x) f 0(x) = lim ∆x→0 ∆x if that limit exists, and if it does we say that f is differentiable at x.

Notation: We will denote this a few different ways. If y = f(x), then all of the following denote the derivative: df d dy f 0(x), , (f(x)), y0, . dx dx dx Examples:

1 1. Find the derivative of f(x) = 2 x + 3 at x = 2. solution: We calculate: f(2 + ∆x) − f(2) f 0(2) = lim ∆x→0 ∆x 1 (2 + ∆x) + 3 − ( 1 2 + 3) = lim 2 2 ∆x→0 ∆x 1 ∆x 1 = 2 = ∆x 2 1 1 So the slope of the tangent line is m = 2 . This makes because y = 2 x + 3 is a line, and the tangent line is the line itself.

2. Find the slope fo the tangent line for g(x) = 1 − x2 at x (unspecified). solution: We have g(x + ∆x) − g(x) g0(x) = lim ∆x→0 ∆x 1 − (x + ∆x)2 − (1 − x2) = lim ∆x→0 ∆x 1 − (x2 + 2x∆x + ∆x2) − (1 − x2) = lim ∆x→0 ∆x −2x∆x − ∆x2 = lim ∆x→0 ∆x = lim (−2x − ∆x) ∆x→0 = −2x

So the slope of the tangent line at x is −2x.

1 3. Find the equation of the tangent line to f(x) = x at the point (1, 1). solution: First we have to find the slope, then we need to use it and the point (1, 1) to find the line’s equation. f(x + ∆x) − f(x) f 0(x) = lim ∆x→0 ∆x 1 − 1 = lim x+∆x x ∆x→0 ∆x x − x+∆x = lim x(x+∆x) x(x+∆x) ∆x→0 ∆x x−(x+∆x) = lim x(x+∆x) ∆x→0 ∆x 1 (−∆x) = lim · ∆x→0 ∆x x(x + ∆x) −1 = lim ∆x→0 x2 + x∆x 1 = − x2 So f 0(1) = −1. Recall that y = mx + b, so here y = −x + b. We sub in the point (1, 1) to find b: 1 = −1 + 2 =⇒ b = 2 =⇒ y = −x + 2.

Notes:

1. As you can see, calculating derivatives from the definition is difficult sometimes. We will develop rules to avoid this.

2. Not every function is differentiable everywhere. In fact, you can’t take the derivative at points

• of discontinuity, • with corners,

• or with vertical .

Note that saying that a function isn’t differentiable at points of discontinuity is equiv- alent to saying that differentiable at x =⇒ continuous at x.