The Lumenary · Grades 9–12

Exponential and logarithmic functions

Name: ______________________

Date: ______________________

You will learn to

Extend exponents to zero, negative and rational values, and prove that the laws of exponents force these definitions
Use logarithms as the inverses of exponential functions, prove the laws of logarithms and solve exponential equations
Model growth and decay with exponential functions, and prove that exponential growth eventually beats every polynomial
Explain the number $e$ as the limit of $(1 + 1/n)^n$ and why $e^{x + y} = e^x e^y$ makes exp an isomorphism from addition to multiplication

1.
Compute $2^{-3} \cdot 2^{5}$ .
2.
Compute $\log_3 81$ .
3.
Compute $16^{3/4}$ .
4.
For $x > 0$ , which expression equals $\log_2(8x)$ ?
- A. $3 + \log_2 x$
- B. $3 \log_2 x$
- C. $8 \log_2 x$
- D. $\log_2 8 \cdot \log_2 x$
5.
Solve $5 \cdot 3^x = 200$ .
- A. $x = \dfrac{\ln 40}{\ln 3} \approx 3.36$
- B. $x = \dfrac{\ln 200}{\ln 15} \approx 1.96$
- C. $x = \dfrac{40}{3}$
- D. $x = \ln\dfrac{40}{3} \approx 2.59$
6.
A hospital receives $160$ mg of iodine-131, which has a half-life of $8$ days. How many milligrams remain after $24$ days?
7.
In the proof that $\log_2 3$ is irrational we reach $2^m = 3^n$ with whole numbers $m \ge 1$ and $n \ge 1$ . Which of these facts would finish the proof? Choose all that apply.
- A. $2^m$ is even but $3^n$ is odd.
- B.By unique prime factorisation, a number greater than $1$ cannot be both a power of $2$ and a power of $3$ .
- C. $3^n$ grows faster than $2^m$ .
- D. $\log_2 3$ lies between $1$ and $2$ .
8.
Let b : ℝ with 0 < b and b ≠ 1. Which Lean statement says that $x \mapsto b^x$ is a bijection from $\mathbb{R}$ onto $(0, \infty)$ ?
- A.∀ y : ℝ, 0 < y → ∃! x : ℝ, b ^ x = y
- B.∀ x : ℝ, ∃! y : ℝ, b ^ x = y
- C.∀ y : ℝ, ∃ x : ℝ, b ^ x = y
- D.∃! x : ℝ, ∀ y : ℝ, 0 < y → b ^ x = y
9.
Savings grow by a rate $r$ per year (for example $r = 0.05$ for $5\%$ ). Write years_to_double(r), which returns the smallest whole number of years $n$ with $(1 + r)^n \ge 2$ , and doubling_time(r), which returns the exact doubling time $\dfrac{\ln 2}{\ln(1 + r)}$ using math.log.
Computational — Practice

Answer key — Exponential and logarithmic functions

1.
$2^{-3} \cdot 2^5 = 2^{-3 + 5} = 2^2 = 4$ . In words: $\tfrac18 \cdot 32 = 4$ .
2.
$3^4 = 81$ , so $\log_3 81 = 4$ .
3.
$16^{3/4} = \left(16^{1/4}\right)^3 = 2^3 = 8$ . (Or $(16^3)^{1/4} = 4096^{1/4} = 8$ : the order does not matter, as the lesson's theorem guarantees.)
4.
$\log_2(8x) = \log_2 8 + \log_2 x = 3 + \log_2 x$ . Check with $x = 4$ : $\log_2 32 = 5 = 3 + 2$ .
5.
$3^x = 40$ , so $x = \log_3 40 = \tfrac{\ln 40}{\ln 3} \approx \tfrac{3.689}{1.099} \approx 3.36$ . Check: $5 \cdot 3^{3.36} \approx 5 \cdot 40 = 200$ .
6.
$A(24) = 160 \cdot \left(\tfrac12\right)^{24/8} = 160 \cdot \tfrac18 = 20$ mg. Each half-life halves the amount: $160 \to 80 \to 40 \to 20$ .
7.
Either parity or unique factorisation finishes the proof: $2^m$ (with $m \ge 1$ ) is even and has only the prime factor $2$ , while $3^n$ is odd and has only the prime factor $3$ . Statements about growth or about the size of $\log_2 3$ cannot exclude a particular equality.
8.
∀ y : ℝ, 0 < y → ∃! x : ℝ, b ^ x = y says each positive $y$ equals $b^x$ for exactly one $x$ ; together with $b^x > 0$ this is the bijection onto $(0, \infty)$ . The order of quantifiers is everything: swapping ∃! and ∀ y gives a false statement.
9.
```
import math

def years_to_double(r):
    n, amount = 0, 1.0
    while amount < 2:
        amount *= 1 + r
        n += 1
    return n

def doubling_time(r):
    return math.log(2) / math.log(1 + r)
```
At $5\%$ the exact doubling time is about $14.2$ years, so the money has doubled at the end of year $15$ . The rule of $70$ gives $70/5 = 14$ , because $\ln(1 + r) \approx r$ for small $r$ .