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Chapter 1 — Exponents and Logarithms

Topics in this chapter:

1.1 Integer Exponents

Exponentiation is shorthand for repeated multiplication, just as multiplication is shorthand for repeated addition.

\[a^n = \underbrace{a \cdot a \cdot a \cdots a}_{n \text{ times}}\]
  • Base: the number being multiplied ($a$)
  • Exponent / Power: the number of times the base is multiplied ($n$)
  • $a^2$ is read “$a$ squared”; $a^3$ is “$a$ cubed.”

Key Rules for Integer Exponents

Rule Formula Example
Product Rule $a^m \cdot a^n = a^{m+n}$ $2^5 \cdot 2^6 = 2^{11} = 2048$
Quotient Rule $\dfrac{a^m}{a^n} = a^{m-n}$ $\dfrac{3^{15}}{3^{12}} = 3^3 = 27$
Power of a Power $(a^m)^n = a^{mn}$ $(3^5)^2 = 3^{10}$
Zero Exponent $a^0 = 1$ (for $a \neq 0$) $7^0 = 1$
Negative Exponent $a^{-n} = \dfrac{1}{a^n}$ $3^{-2} = \frac{1}{9}$

Warning: These rules only apply when the bases are the same. You cannot combine $2^3 \cdot 3^4$ using the product rule.

Example 1-1: What is $2^5 \cdot 2^6$?

Solution: $2^5 \cdot 2^6 = 2^{5+6} = 2^{11} = 2048$.

Example 1-2: What is $\dfrac{3^{15}}{3^{12}}$?

Solution: $\dfrac{3^{15}}{3^{12}} = 3^{15-12} = 3^3 = 27$.

Example 1-3: What is the difference between $\dfrac{5^5 + 5^2}{5}$ and $\dfrac{(5^5)(5^2)}{5}$?

Solution: In the first expression, $5$ must divide each term of the numerator:

\[\frac{5^5 + 5^2}{5} = \frac{5^5}{5} + \frac{5^2}{5} = 5^4 + 5\]

In the second, the $5$ in the denominator cancels with one factor:

\[\frac{(5^5)(5^2)}{5} = \frac{5^7}{5} = 5^6\]

Key Distinction: $\dfrac{a + b}{c} = \dfrac{a}{c} + \dfrac{b}{c}$ (divide into each term), but $\dfrac{ab}{c}$ only requires dividing one factor. This matters for expressions like $\dfrac{3+x}{3}$ (not reducible) vs. $\dfrac{3x}{3} = x$.

Example 1-4: Evaluate $(3^5)^2$ and $(4^{-3})^{-2}$.

Solution:

  • $(3^5)^2 = 3^{5 \cdot 2} = 3^{10}$
  • $(4^{-3})^{-2} = 4^{(-3)(-2)} = 4^6$

Example 1-5: What is the difference between $2^{3^4}$ and $(2^3)^4$?

Solution:

  • $2^{3^4} = 2^{81}$ (the exponent itself is raised to the 4th power first)
  • $(2^3)^4 = 2^{3 \cdot 4} = 2^{12}$

These are very different values!

Warning about $0^0$: We cannot determine $0^0$ using the rule $a^0 = a^n/a^n = 1$, because $0^n/0^n$ has $0$ in the denominator. Thus $0^0$ is undefined. Also, $0$ raised to any negative power is undefined, but $0$ raised to any positive power is $0$.

Exercise 1-1: Evaluate each of the following.

  1. $3^4$
  2. $2^5 \cdot 2^2$
  3. $5^{-3} \cdot 5^5 \cdot 5^{-1}$
  4. $4^3 / 4$
  5. $2^7 / 2^2$
  6. $(3^4 \cdot 3^{-2})/(3^5 \cdot 3^{-2})$
  7. $2^5 \cdot 3^2 \cdot 2^{-3}$
  8. $5^2 \cdot 3^{-1} \cdot 2 \cdot 4 \cdot 5^{-1} \cdot 2^{-2}$

1.2 Fractional Exponents

Fractional exponents extend the idea of exponents to non-integers:

\[a^{1/n} = \sqrt[n]{a}\]

The exponent $\frac{1}{2}$ is the square root, $\frac{1}{3}$ is the cube root, and in general $\frac{1}{n}$ is the $n$th root.

For general fractions:

\[a^{m/n} = \left(a^{1/n}\right)^m = \left(\sqrt[n]{a}\right)^m\]

Example 1-6: Which two integers is $\sqrt{55}$ between?

Solution: Since $\sqrt{49} = 7$ and $\sqrt{64} = 8$, and $49 < 55 < 64$, we know $7 < \sqrt{55} < 8$.

Example 1-7: Evaluate the following:

  1. $8^{5/3} = (8^{1/3})^5 = 2^5 = 32$
  2. $(\sqrt{8})^{2/3} = (8^{1/2})^{2/3} = 8^{1/3} = 2$
  3. $\sqrt[4]{81^{-3}} = 81^{-3/4} = (81^{1/4})^{-3} = 3^{-3} = \frac{1}{27}$
  4. $\left(\frac{1}{8}\right)^{2/3} = \frac{1^{2/3}}{8^{2/3}} = \frac{1}{4}$

Example 1-8: Evaluate $\dfrac{(4^{2/3})(2^{1/6})(3^{3/2})}{(2^{-1/2})(3^{1/2})}$.

Solution: Convert $4^{2/3} = (2^2)^{2/3} = 2^{4/3}$. Then:

\[\frac{2^{4/3} \cdot 2^{1/6} \cdot 3^{3/2}}{2^{-1/2} \cdot 3^{1/2}} = 2^{4/3 + 1/6 - (-1/2)} \cdot 3^{3/2 - 1/2} = 2^{4/3 + 1/6 + 1/2} \cdot 3^1 = 2^2 \cdot 3 = 12\]

Warning — Positive vs. Negative Roots:

  • If the radical sign or fractional power is already in the problem, give only the positive root.
  • If you introduce the fractional power yourself (e.g., $x^2 = 9$), give all solutions: $x = \pm 3$.
  • For odd roots, there is no ambiguity: $\sqrt[3]{-8} = -2$.

Example 1-9: What is the difference between $x^2 = 9$ and $x = \sqrt{9}$?

Solution: In $x^2 = 9$, we answer $x = \pm 3$. In $x = \sqrt{9}$, the radical is already present, so $x = 3$ (positive only).

Exercise 1-2: Try these.

  1. $9^{3/2}$
  2. $(\sqrt[4]{81})^{3/2}$
  3. $64^{-4/3}$
  4. $\sqrt[5]{100000^3}$
  5. $(4/9)^{-3/2}$
  6. $(\sqrt[4]{1/16})^{-3}$

Exercise 1-3: Find all real $x$ in each of the following.

  1. $x = (-2)^5$
  2. $x = \sqrt[3]{1/8}$
  3. $x^6 = 64$
  4. $x^3 = 64$
  5. $x = (-27)^{-2/3}$
  6. $x^{5/3} = 243$

1.3 Simplifying Radical Expressions

To simplify a radical expression, remove all factors that can come out from under the radical:

  1. Write the prime factorization of the radicand.
  2. Convert to exponential notation: $\sqrt[n]{a^k} = a^{k/n}$.
  3. Separate each factor: $(a^j \cdot b^k)^{1/n} = a^{j/n} \cdot b^{k/n}$.
  4. Extract factors with integer exponents (those where the exponent $\ge 1$).
  5. Combine remaining fractional-exponent factors under a single radical.

Example: Simplify $\sqrt{8}$.

\[\sqrt{8} = (4 \cdot 2)^{1/2} = 4^{1/2} \cdot 2^{1/2} = 2\sqrt{2}\]

By inspection: $8 = 4 \times 2$, and $\sqrt{4} = 2$, so $\sqrt{8} = 2\sqrt{2}$.

Example 1-11: Simplify $\sqrt{1440}$, $\dfrac{\sqrt[3]{144}}{\sqrt[3]{125}}$, and $\sqrt[6]{6912}$.

Solution:

$\sqrt{1440}$: The prime factorization of $1440 = 2^5 \cdot 3^2 \cdot 5$.

\[\sqrt{1440} = (2^5 \cdot 3^2 \cdot 5)^{1/2} = 2^{5/2} \cdot 3 \cdot 5^{1/2} = 4 \cdot 3 \cdot \sqrt{2 \cdot 5} = 12\sqrt{10}\]

By inspection: $1440 = 144 \times 10$, and $\sqrt{144} = 12$, so $\sqrt{1440} = 12\sqrt{10}$.

$\dfrac{\sqrt[3]{144}}{\sqrt[3]{125}}$: $\quad \dfrac{(2^4 \cdot 3^2)^{1/3}}{(5^3)^{1/3}} = \dfrac{2^{4/3} \cdot 3^{2/3}}{5} = \dfrac{2 \cdot (2^{1/3} \cdot 3^{2/3})}{5} = \dfrac{2\sqrt[3]{12}}{5}$

$\sqrt[6]{6912}$: $\quad 6912 = 2^8 \cdot 3^3$

\[\sqrt[6]{6912} = 2^{8/6} \cdot 3^{3/6} = 2^{4/3} \cdot 3^{1/2} = 2 \cdot (2^{1/3} \cdot 3^{1/2}) = 2\sqrt[6]{108}\]

Exercise 1-4: Simplify the following.

  1. $\sqrt{27}$
  2. $\sqrt[3]{128}$
  3. $\sqrt[4]{1600}$
  4. $\sqrt{9095625}$
  5. $\dfrac{\sqrt[3]{36000}}{243}$
  6. $\dfrac{56}{\sqrt{126}}$

1.4 Rationalizing Denominators

Rationalizing the denominator means rewriting a fraction so that the denominator contains no radicals (irrational numbers). This is done by multiplying numerator and denominator by a suitable expression.

For Simple Radicals

Multiply top and bottom by the radical that makes the denominator rational:

\[\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\]

For Higher Roots

Make the exponent of each factor under the radical equal to the root:

\[\frac{1}{\sqrt[3]{12}} = \frac{1}{\sqrt[3]{2^2 \cdot 3}} \quad\Longrightarrow\quad \text{multiply by } \frac{\sqrt[3]{2 \cdot 3^2}}{\sqrt[3]{2 \cdot 3^2}} = \frac{\sqrt[3]{18}}{6}\]

Conjugate Radicals

For expressions like $\dfrac{1}{1 + \sqrt{2}}$, multiply by the conjugate $1 - \sqrt{2}$:

\[\frac{1}{1+\sqrt{2}} \cdot \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{1-\sqrt{2}}{1 - 2} = \sqrt{2} - 1\]

Conjugate Radical: The conjugate of $a\sqrt{b} + c\sqrt{d}$ is $a\sqrt{b} - c\sqrt{d}$. Their product is always rational:

\[(a\sqrt{b} + c\sqrt{d})(a\sqrt{b} - c\sqrt{d}) = a^2b - c^2d\]

Example 1-12: Rationalize $\dfrac{2\sqrt{5}}{3\sqrt[4]{72}}$.

Solution: Split into primes: $72 = 2^3 \cdot 3^2$. Then:

\[\frac{2\sqrt{5}}{3\sqrt[4]{2^3 \cdot 3^2}}\]

Multiply by $\dfrac{\sqrt[4]{2 \cdot 3^2}}{\sqrt[4]{2 \cdot 3^2}}$ to get $\dfrac{\sqrt[4]{450}}{9}$.

Example 1-14: Rationalize $\dfrac{1}{5 - \sqrt{2} - \sqrt{3}}$.

Solution: Treat as $(5 - \sqrt{2}) - \sqrt{3}$ and multiply by $(5-\sqrt{2}) + \sqrt{3}$. After simplification and a second rationalization step:

\[\frac{13 + 25\sqrt{2} + 5\sqrt{3} + 6\sqrt{6}}{94}\]

Exercise 1-5: Rationalize the denominators.

  1. $3/\sqrt{3}$
  2. $\sqrt{2}/\sqrt{6}$
  3. $2/\sqrt[3]{24}$
  4. $1/\sqrt[4]{1800}$
  5. $5^{1/3}/5^{5/3}$
  6. $(3a^2 \cdot 2^{2/3})/(3^{1/6} \cdot 2^{3/2})$

Exercise 1-6: Rationalize the denominators.

  1. $\dfrac{1}{\sqrt{7} + \sqrt{3}}$
  2. $\dfrac{6}{\sqrt{15} - \sqrt{6}}$
  3. $\dfrac{\sqrt{2}}{\sqrt{6} - 2}$
  4. $\dfrac{1}{1 - \sqrt{2} + \sqrt{3}}$ (Hint: Multiply by $1 + \sqrt{2}$ first.)

1.5 Logarithms

Definition: $\log_b a = c$ means $b^c = a$.

Read: “The logarithm of $a$ base $b$ is $c$” — meaning “the power you raise $b$ to in order to get $a$.”

Exponential Form Logarithmic Form
$2^3 = 8$ $\log_2 8 = 3$
$3^5 = 243$ $\log_3 243 = 5$
$5^2 = 25$ $\log_5 25 = 2$

Key Logarithm Properties

Property Formula
Product Rule $\log_b(MN) = \log_b M + \log_b N$
Quotient Rule $\log_b(M/N) = \log_b M - \log_b N$
Power Rule $\log_b(M^k) = k \log_b M$
Change of Base $\log_b a = \dfrac{\log_c a}{\log_c b}$
Identity $\log_b b = 1$, $\quad \log_b 1 = 0$

Warning: The base and the argument of the logarithm must both be positive for the result to be real. When $\log$ is written without a base, it is usually assumed to be base 10.

Example 1-15: Evaluate $\log_3 \frac{1}{243}$ and $\log_8 2$.

Solution:

  • $\log_3 \frac{1}{243}$: We need $3^x = \frac{1}{243} = 3^{-5}$, so $x = -5$.
  • $\log_8 2$: We need $8^x = 2$. Since $8 = 2^3$: $(2^3)^x = 2^1 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}$.

Exercise 1-7: Convert to logarithmic form.

  1. $3^3 = 27$
  2. $16^{1/4} = 2$
  3. $x^2 = y$

Exercise 1-8: Convert to exponential form.

  1. $\log_2 8 = 3$
  2. $\log_3(1/9) = -2$
  3. $\log_x y = z$

Exercise 1-9: Find each of the following.

  1. $\log_5 625$
  2. $\log_{1/2} 2$
  3. $\log_9 \sqrt{3}$
  4. $\log_{\sqrt[3]{5}} \sqrt{5}$

The Big Picture — Why Logarithms Matter

Logarithms were first invented by John Napier in the early 1600s as a trick to do multiplication — since $\log(MN) = \log M + \log N$, two large numbers could be multiplied by simply adding their logarithms from a table.

Other uses:

  • Slide rules work on the same principle.
  • Logarithmic scales (like the decibel for sound) compress huge ranges into manageable numbers. A whisper is ~20 dB; a jet engine is ~120 dB — meaning a jet is $10^{10}$ (10 billion) times louder in power, not “6 times” as sometimes misreported.
  • Logarithmic graph paper turns exponential curves into straight lines.

Exercises & Problems

Problems to Solve for Chapter 1:

  1. Find the value of $\log_5 (125\sqrt[6]{5} \cdot 25)$. (AHSME 1950)
  2. What is the logarithm of $27\sqrt[3]{9}\sqrt{9}$ base 3? (AHSME 1953)
  3. Express $2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}$ in simplest form. (AHSME 1958)
  4. Find $(-3)^{-2} + (-2)^{-1} + (-1)^0 + 0^1 + 1^2 + 2^3 + 3^4$.
  5. Simplify the expression $81^{-1/2} \cdot 2^{2x}$. (AHSME 1965)
  6. Find, with a rational common denominator, $\dfrac{1}{\sqrt{1}+\sqrt{2}} + \dfrac{1}{\sqrt{2}+\sqrt{3}} + \cdots + \dfrac{1}{\sqrt{n}+\sqrt{n+1}}$.
  7. Write $\dfrac{1}{\sqrt{2} + \sqrt{3} - \sqrt{5}}$ with a rational denominator. (AHSME 1952)
  8. Find $\log_3 \sqrt[3]{9}$.
  9. Solve for $n$: $\sqrt{1 + \sqrt{1 + \sqrt{n}}} = 2$. (MATHCOUNTS 1991)
  10. Find $x$ if $216^x = 16^{2x}$. (Mandelbrot #3)
  11. Solve the equation $\log_{2x} 216 = x$, where $x$ is real. (AHSME 1960)
  12. Suppose $A$ and $B$ are positive reals with $\log_A B = \log_B A$. If neither is 1 and $A \neq B$, find $AB$. (MAG 1992)
  13. The formula $N = 8 \times 10^8 \cdot x^{-3/2}$ gives the number of individuals whose income exceeds $x$ dollars. Find the smallest possible lowest income of the wealthiest 800 individuals. (AHSME 1960)
  14. Show that if $a \neq c$, $a^x = c^q$, and $c^y = a^z$, then $xy = qz$. (AHSME 1951)
  15. Given $\log_3 2 = 0.631$, find the smallest positive integer $a$ such that $3^a > 2^{102}$. (Mandelbrot #3)
  16. Show that $\log_6 2 + \log_6 3 = 1$.

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