Calculus II ep10: Other logarithms (Sep 30, 2026)

Overview

Prof Staecker develops differentiation and integration rules for exponentials and logarithms with bases other than e, including the change-of-base formula and the derivative of log base b. He applies logarithms to a hemochromatosis treatment example: reducing ferritin from 800 to 20 by removing 10% per phlebotomy is modeled by 800(0.9)^t, which reaches the target after about 35 treatments.

Key takeaways

For b > 0 and b ≠ 1, differentiating b^x multiplies the original exponential by ln(b), while integrating it divides by ln(b).
A logarithm is an inverse exponent operation: log_b(y) equals the exponent x that solves b^x = y.
The change-of-base identity log_b(x) = ln(x)/ln(b) lets a calculator evaluate logarithms even when it has no dedicated button for base b.
The derivative of log_b(x) is 1/[x ln(b)]; the familiar natural-log derivative 1/x is the special case b = e.
Repeatedly retaining 90% of a quantity produces an exponential model: starting at 800, the sequence is 800(0.9)^t, not a constant decrement.
Solving 800(0.9)^t = 20 with logarithms yields approximately 35.01 reductions, demonstrating how logs isolate an unknown exponent.

Chapters

0:00 Integrating b^x and Distinguishing Exponentials from Power Functions

For a positive base b other than 1, the derivative of b^x is b^x ln(b), so its antiderivative is b^x/ln(b) + C.
The power rule applies to x^n when x is the base; it does not apply when x appears in the exponent, as in 3^x.
For b^(kx), integration requires division by both k and ln(b): b^(kx)/(k ln(b)) + C.

10:10 Using Chain Rule and Exponential Antiderivative Formulas

For a function such as 4.2^(3x^2 - 7), differentiate the exponential, multiply by ln(4.2), then apply the chain rule factor 6x.
The natural-base formula remains simpler: the antiderivative of e^x is e^x + C, while that of e^(kx) is e^(kx)/k + C.
The worked exercises contrast ordinary power functions, such as x^4, with exponentials whose variable is in the exponent.

13:10 Defining Logarithms as Inverses of Exponentials

The equation y = b^x is equivalent to x = log_b(y); a logarithm asks which exponent on b produces its argument.
Prof Staecker illustrates the definition with log_2(8) = 3 because 2^3 = 8.
A logarithm such as log_2(7) generally cannot be found exactly by inspection and requires a calculator approximation.

20:00 Evaluating Logarithms with Negative and Fractional Exponents

Since 4^(-2) = 1/16, log_4(1/16) = -2; reciprocals of powers can correspond to negative logarithm values.
For any valid base b, log_b(1) = 0 because b^0 = 1.
Because 9^(1/2) = 3, log_9(3) = 1/2; by contrast, log_9(18) is not a simple whole-number exponent.

25:30 Common Logarithms and Base-10 Magnitude

In common notation, log without a written base means log base 10; ln denotes the natural logarithm, or log base e.
For example, log(1,000) = 3 and log(10,000) = 4 because these numbers are 10^3 and 10^4.
A three-digit positive number lies between 10^2 and 10^3, so its base-10 logarithm lies between 2 and 3; digit count gives a rough scale, not an exact value.

31:00 Deriving the Change-of-Base Formula for Calculator Use

Starting with y = log_b(x), rewrite it as b^y = x, take natural logarithms, and use ln(b^y) = y ln(b).
Solving for y gives log_b(x) = ln(x)/ln(b), allowing a calculator's ln button to evaluate logs with other bases.
For example, log_2(6767) can be entered as ln(6767)/ln(2); the formula changes the logarithm's base without changing its value.

34:30 Differentiating Logarithms with Non-Natural Bases

Substitute log_b(x) = ln(x)/ln(b); because ln(b) is constant with respect to x, it factors out of the derivative.
The resulting rule is d/dx[log_b(x)] = 1/[x ln(b)], which reduces to 1/x when b = e.
Prof Staecker distinguishes this derivative rule from integrating ln(x), an antiderivative topic reserved for later.

38:00 Hemochromatosis, Ferritin, and Phlebotomy

Prof Staecker describes his hemochromatosis diagnosis, a genetic condition that can cause excess iron to accumulate in the body.
In his example, ferritin measured about 800 at diagnosis, compared with a target near 20; the treatment is phlebotomy, or removing blood.
Each treatment removes about 10% of the remaining iron in the model, so the amount lost decreases over successive sessions rather than staying constant.

44:00 Modeling Repeated 10% Reductions with an Exponential Function

Keeping 90% of the ferritin level after each phlebotomy gives f(1) = 800(0.9) and f(2) = 800(0.9)^2.
After t treatments, the model is f(t) = 800(0.9)^t; setting f(t) = 20 asks when the level reaches the target.
The model represents compounding percentage reductions, not a fixed amount subtracted at every appointment.

47:30 Solving for About 35 Phlebotomy Treatments with Logarithms

From 800(0.9)^t = 20, divide by 800 to obtain (0.9)^t = 1/40.
Taking natural logs gives t ln(0.9) = ln(1/40), so t = ln(1/40)/ln(0.9).
The calculator result is about 35.01 treatments, close to the doctor's initial plan of phlebotomy every two weeks for a year.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Prof Staecker.