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3 Indices, Surds and Logarithms

3.1 Indices (Exponents) and Surds

Definition of Indices

When discussing indices (or exponents), a positive integer exponent is defined as:
an=aimesaimesaimesext…(nfactors)a^n = a imes a imes a imes ext{ … (n factors)}
where aa is called the base, and nn is the index or exponent.

Example of Exponentiation

For example, the expression for 545^4 can be expressed as:
54=5imes5imes5imes55^4 = 5 imes 5 imes 5 imes 5.

Restrictions on Bases
  • The discussion will focus on positive bases (i.e., a>0a > 0).
  • The definitions can be extended to cover zero, negative, and fractional indices.
Extended Definitions for Indices

For a>0a > 0 and positive integers pp and qq, we have the following relationships:

  • a0=1a^0 = 1,
  • a−p=1apa^{-p} = \frac{1}{a^p},
  • a^{ rac{1}{n}} = ext{ }
    oot{n}{a} .
Specific Cases for Indices
  • Example: 20=12^0 = 1,
  • Example: 2−3=123=182^{-3} = \frac{1}{2^3} = \frac{1}{8}.
  • The result apextforpextbeinganintegerandaextbeingapositivenumbera^p ext{ for } p ext{ being an integer and } a ext{ being a positive number} applies across various contexts, such as roots: extForexample,42=(212)3=23ext{For example, } 4^2 = (2^{\frac{1}{2}})^3 = 2^3.
Rules of Indices

The following rules apply for positive bases aa and bb with any rational indices mm and nn:

  • amimesan=am+na^m imes a^n = a^{m+n} (rule for the same base),
  • amimesbm=(ab)ma^m imes b^m = (ab)^m,
  • am/an=am−na^m \big/ a^n = a^{m-n},
  • (am)n=amn(a^m)^n = a^{mn}.
Example Problems
  1. Simplifying the following expressions:
    • (a) 93imes96=93+6=999^3 imes 9^6 = 9^{3+6} = 9^9
    • (b) (94)3=94imes3=912( 9^4)^3 = 9^{4 imes 3} = 9^{12}
    • (c) 12363=123(2imes3)3=12323imes33\frac{12^3}{6^3} = \frac{12^3}{(2 imes 3)^3} = \frac{12^3}{2^3 imes 3^3}
    • (d) 232imes213=232+132^\frac{3}{2} imes 2^{\frac{1}{3}} = 2^{\frac{3}{2} + \frac{1}{3}}=
    • Solving numeric conversions gives roots essential to understand further properties.

3.2 Exponential Equations

Definition

An equation that contains a variable in the exponent is termed an exponential equation, expressed typically as a=ba = b.
If bb can be expressed as ana^n, then the relationship arises as follows:
a=anextleadstox=nextassuminga<br/>≠0,1a = a^n ext{ leads to } x=n ext{ assuming } a <br />\neq 0, 1.

Example Solutions of Exponential Equations
  • Example 5: Solve the equation 4x=164^{x} = 16.

    1. Recognizing that 16=4216 = 4^{2}, then equate: \$ x = 2 \$.
  • Example 6: Solve 5−2=1255^{-2} = \frac{1}{25}

    1. Rewriting gives 5−2=5−25^{-2} = 5^{-2}, solutions yield x outcomes easily confirmed.

3.3 Logarithms

Basic Definitions and Relationships

From previous knowledge, we conclude:

  • The logarithmic function connects directly back to exponential states: if ax=ba^x = b,
    then x=extloga(b)x = ext{log}_a(b)

  • It’s essential for logarithms (base > 0 and not equal to 1) where:
    a>0,a<br/>≠1extleadstoy>0a > 0, a <br />\neq 1 ext{ leads to } y > 0.

Converting Between Forms
  • Changing to logarithmic from index forms:%
    • Example 9: Change a common base form into logarithmic:
    1. 42=164^2 = 16 translates to 2=extlog4162 = ext{log}_4 16,
    2. The logarithms aid in representing equivalences across mathematical structures.
Properties of Logarithms
  1. Product Law: extloga(xy)=extloga(x)+extloga(y)ext{log}_a(xy) = ext{log}_a(x) + ext{log}_a(y).
  2. Quotient Law: extloga(xy)=extloga(x)−extloga(y)ext{log}_a \bigg(\frac{x}{y}\bigg) = ext{log}_a(x) - ext{log}_a(y).
  3. Power Law: extloga(xn)=nextloga(x)ext{log}_a(x^n) = n ext{log}_a(x).

These properties facilitate simplification and concurrency in algebraic formulations. Each law enables breakdown and recompilation of function bases within calculus and algebra itself.

3.4 Common and Natural Logarithms

Common Logarithms

Logarithms using base 10 are termed common, frequently noted with logs:

  • Notably, conversion gives log1010=1log_{10} 10 = 1.
Natural Logarithms

Natural logarithms are confirmed for base e (approximately 2.71828). Using:

  • Identified notation becomes extlnextx=extloge(extx)ext{ln } ext{x} = ext{log}_e( ext{x}), appropriately linking numerics and sequences clearly through scientific bases.
Application Example

Common logarithms resolve through transformation consistently. For example, evaluating x+2=21x + 2 = 21 helps rearrange and solve accordingly.