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maths1-week6

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Deep study for Quiz 2 week 6. Logarithm is the inverse operation of exponentiation — every log question hides an exponent question.

Week map

Log as inverse of exp → definition and domain → log laws → change of base → solve log equations → check solutions.

Logarithm notation

$\log_a x$ → “log base $a$ of $x$ ” → the exponent $y$ such that $a^y = x$ → requires $a > 0$ , $a \neq 1$ , $x > 0$ .
$\ln x$ → natural log → base $e$ → $\ln x = \log_e x$ .
$\log_{10} x$ or $\log x$ → common log (context-dependent).

Mini-example:

\log_2 8 = 3

because

2^3 = 8

\log_5 1 = 0

because

5^0 = 1

\log_3 3 = 1

Key identities

$a^{\log_a x} = x$ for $x > 0$ .
$\log_a(a^x) = x$ for all real $x$ .
$\log_a 1 = 0$ ; $\log_a a = 1$ .

Log laws

Product: $\log_a(xy) = \log_a x + \log_a y$ .
Quotient: $\log_a(x/y) = \log_a x - \log_a y$ .
Power: $\log_a(x^n) = n \log_a x$ .
Change of base: $\log_a x = \frac{\log_b x}{\log_b a}$ for any valid base $b$ .

Mini-example:

\log_2(16 \cdot 4) = \log_2 16 + \log_2 4 = 4 + 2 = 6

. Check:

2^6 = 64 = 16 \cdot 4

Graph features

Domain $(0, \infty)$ ; vertical asymptote at $x = 0$ .
Passes through $(1, 0)$ and $(a, 1)$ .
Reflection of $y = a^x$ over $y = x$ .

Pattern families

Easy — Evaluate simple logs

Rewrite as “what power of base gives this number?” Use

\log_a 1 = 0

\log_a a = 1

. Connect base-10 and natural logs to familiar values.

Medium — Apply log laws

Expand or condense expressions. Split products into sums; bring exponents down. Change base to evaluate unfamiliar logs.

Hard — Solve log and exponential equations

Isolate log or exponent; convert between forms. Always check solutions in original equation — extraneous values appear when domain violated.

Worked mini-examples

Example 1 — Definition.

\log_3 81 = ?

→

3^y = 81 = 3^4

→ answer 4.

Example 2 — Product law.

\log_{10} 50 + \log_{10} 2 = \log_{10} 100 = 2

Example 3 — Power law.

\log_2(x^5) = 5 \log_2 x

Example 4 — Change of base.

\log_4 8 = \frac{\ln 8}{\ln 4} = \frac{3\ln 2}{2\ln 2} = 3/2

Example 5 — Solve.

\log_2(x - 1) = 3

→

x - 1 = 2^3 = 8

→

x = 9

. Check:

\log_2 8 = 3

✓. Domain requires

x - 1 > 0

, so

x > 1

— satisfied.

Traps

Taking $\log$ of negative number or zero — undefined.
Using $\log(x + y) = \log x + \log y$ — false; product law only.
Forgetting $0!$ -style edge: $\log_a 1 = 0$ , not undefined.
Dividing logs incorrectly: $\log x / \log y \neq \log(x/y)$ .
Accepting solution that makes log argument $\leq 0$ .

Diagnostic (try yourself)

Evaluate $\log_5 125$ and $\log_3(1/9)$ .
Expand $\log_2(8x^3)$ using log laws.
Use change of base to express $\log_8 4$ as a fraction.
Solve $\log_{10}(x) = 2$ . State the domain condition.
True or false: $\ln(e^5) = 5$ ? What is $\ln 1$ ?