12.1 Logarithms (Logaritma), Complex Numbers & Arithmetic/Geometric Sequences (Diziler)

12.1 Logarithms (Logaritma), Complex Numbers ($i^2 = -1$) & Sequences (Diziler)

This section extends independent mathematics practice with logarithms and complex numbers, then develops arithmetic and geometric sequences. The current official guide does not enumerate these as guaranteed subtopics. Use the extensions after the basic arithmetic, algebra and released-question work rather than assigning them an invented item frequency.

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1. Logarithms (Logaritma)

Definition & Strict Domain Conditions (Tanım Kümesi)

The logarithm is the inverse operation of exponentiation. For a base $a$ and argument $f(x)$:

$$y = \log_a(f(x)) \iff a^y = f(x)$$

Whenever a TR-YÖS problem asks for the largest domain (en geniş tanım kümesi) of $y = \log_{g(x)}(f(x))$, or when you solve logarithmic equations and inequalities, you must enforce three strict conditions:

1. Positive argument: $f(x) > 0$
2. Positive base: $g(x) > 0$
3. Base not equal to 1: $g(x) \ne 1$

Two standard shorthand notations appear constantly:

• Common logarithm (Bayağı Logaritma): $\log x = \log_{10} x$ (base $10$)
• Natural logarithm (Doğal Logaritma): $\ln x = \log_e x$ (base $e \approx 2.718$)

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Core Logarithmic Identities (Logaritma Özellikleri)

Assuming $a > 0, a \ne 1$ and $x, y > 0$:

• Unit & Zero Rules:

$$\log_a 1 = 0, \qquad \log_a a = 1, \qquad \ln e = 1, \qquad \log 10 = 1$$

• Product & Quotient Rules (Çarpım ve Bölüm Kuralları):

$$\log_a(x \cdot y) = \log_a x + \log_a y, \qquad \log_a\!\left(\frac{x}{y}\right) = \log_a x - \log_a y$$

(Warning: $\log_a(x + y) \ne \log_a x + \log_a y$! Sums inside a logarithm cannot be split.)

• Power Extraction Rule (Kuvvet Kuralı):

$$\log_{a^n}(b^m) = \frac{m}{n}\log_a b$$

For example, $\log_{8}(16) = \log_{2^3}(2^4) = \frac{4}{3}\log_2 2 = \frac{4}{3}$, and $\log_{\sqrt{a}}(a^3) = \log_{a^{1/2}}(a^3) = \frac{3}{1/2} = 6$.

• Exponent-Swap & Cancellation Rules:

$$a^{\log_b c} = c^{\log_b a}, \qquad a^{\log_a x} = x$$

For instance, $4^{\log_3 5} = 5^{\log_3 4}$ and $e^{\ln(x+2)} = x + 2$ (for $x > -2$).

Change of Base & Telescoping Chain Product (Taban Değiştirme)

To convert a logarithm from base $a$ to any convenient new base $c$ (where $c > 0, c \ne 1$):

$$\log_a b = \frac{\log_c b}{\log_c a} = \frac{\ln b}{\ln a}$$

Two useful change-of-base corollaries are:

1. Reciprocal Base-Argument Flip:

$$\log_a b = \frac{1}{\log_b a} \iff \log_a b \cdot \log_b a = 1$$

When an equation contains both $\log_a x$ and $\log_x a$, substitute $t = \log_a x$ and $\frac{1}{t} = \log_x a$ to turn the equation into a quadratic in $t$. 2. Telescoping Chain Product:

$$\log_a b \cdot \log_b c \cdot \log_c d = \frac{\ln b}{\ln a} \cdot \frac{\ln c}{\ln b} \cdot \frac{\ln d}{\ln c} = \frac{\ln d}{\ln a} = \log_a d$$

Logarithmic Inequalities (Logaritmik Eşitsizlikler)

When solving $\log_a f(x) < \log_a g(x)$ (or $\log_a f(x) < k$):

• Case 1 ($a > 1$, increasing function): Preserve the inequality direction: $0 < f(x) < g(x)$.
• Case 2 ($0 < a < 1$, decreasing function): Reverse the inequality direction, and never forget the positivity requirement $f(x) > 0$: $f(x) > g(x) > 0$.

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2. Complex Numbers (Karmaşık Sayılar — $i^2 = -1$)

The imaginary unit $i$ is defined by $i = \sqrt{-1}$, so $i^2 = -1$. Every complex number is written in standard form as $z = a + bi$ ($a, b \in \mathbb{R}$), where $\operatorname{Re}(z) = a$ is the real part (gerçel kısım) and $\operatorname{Im}(z) = b$ is the imaginary part (sanal kısım). Two complex numbers are equal if and only if both their real parts and their imaginary parts match.

Period-4 Cycle of Powers of $i$

Because $i^4 = (i^2)^2 = (-1)^2 = 1$, integer powers of $i$ repeat every $4$ steps according to the remainder of the exponent modulo $4$:

$$i^{4k} = 1, \qquad i^{4k+1} = i, \qquad i^{4k+2} = -1, \qquad i^{4k+3} = -i$$

For negative exponents, add a multiple of $4$ until the exponent is positive (e.g., $i^{-13} = i^{-13 + 16} = i^3 = -i$). Crucially, any four consecutive powers of $i$ sum to zero:

$$i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$$

High-Speed Binomial & Quotient Identities

Instead of expanding large powers of $1 \pm i$ by hand, memorize these four instant shortcuts:

$$(1 + i)^2 = 1 + 2i + i^2 = 2i, \qquad (1 - i)^2 = 1 - 2i + i^2 = -2i$$ $$(1 + i)(1 - i) = 1 - i^2 = 2, \qquad \frac{1 + i}{1 - i} = i, \qquad \frac{1 - i}{1 + i} = -i$$

Thus, to compute $(1 + i)^{12}$, rewrite it as $[(1 + i)^2]^6 = (2i)^6 = 2^6 \cdot i^6 = 64 \cdot (-1) = -64$.

Conjugate (Eşlenik) & Modulus (Modül / Mutlak Değer)

For $z = a + bi$, its complex conjugate is $\bar{z} = a - bi$ (only the sign of $i$ changes!), and its modulus is its distance from the origin in the complex plane:

$$\lvert z \rvert = \sqrt{a^2 + b^2}, \qquad z \cdot \bar{z} = (a + bi)(a - bi) = a^2 + b^2 = \lvert z \rvert^2$$

Modulus respects multiplication, division, and powers: $\lvert z_1 \cdot z_2 \rvert = \lvert z_1 \rvert \cdot \lvert z_2 \rvert$, $\left\lvert \frac{z_1}{z_2} \right\rvert = \frac{\lvert z_1 \rvert}{\lvert z_2 \rvert}$, and $\lvert z^n \rvert = \lvert z \rvert^n$.

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3. Arithmetic & Geometric Sequences (Aritmetik ve Geometrik Diziler)

A sequence (dizi) is a function $a_n$ whose domain is the positive integers $\mathbb{Z}^+ = \{1, 2, 3, \dots\}$.

Arithmetic Sequence (Aritmetik Dizi)

Consecutive terms differ by a constant common difference (ortak fark) $d = a_{n+1} - a_n$:

• General Term Relationship: $a_n = a_1 + (n - 1)d$, and more generally between any two terms $a_n$ and $a_p$:

$$a_n = a_p + (n - p)d \implies d = \frac{a_n - a_p}{n - p}$$

• Equidistant Symmetry (Ortanca Terim Özelliği): Every term is the arithmetic mean of terms equidistant from it:

$$a_k = \frac{a_{k-m} + a_{k+m}}{2} \quad \text{(and if } n + p = q + r, \text{ then } a_n + a_p = a_q + a_r\text{)}$$

• Sum of First $n$ Terms (İlk n Terim Toplamı):

$$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}\bigl[2a_1 + (n - 1)d\bigr]$$

Geometric Sequence (Geometrik Dizi)

Consecutive terms have a constant common ratio (ortak çarpan) $r = \frac{a_{n+1}}{a_n}$:

• General Term Relationship: $a_n = a_1 \cdot r^{n-1}$, and between any two indices $n$ and $p$:

$$a_n = a_p \cdot r^{n-p} \implies r^{n-p} = \frac{a_n}{a_p}$$

• Equidistant Symmetry: The square of any term equals the product of terms equidistant from it:

$$a_k^2 = a_{k-m} \cdot a_{k+m} \quad \text{(and if } n + p = q + r, \text{ then } a_n \cdot a_p = a_q \cdot a_r\text{)}$$

• Finite & Infinite Geometric Sums:

$$S_n = a_1 \frac{1 - r^n}{1 - r} \quad (r \ne 1), \qquad S_\infty = \sum_{n=1}^{\infty} a_1 r^{n-1} = \frac{a_1}{1 - r} \quad (\lvert r \rvert < 1)$$

Recovering $a_n$ from Partial Sums $S_n$

For any sequence (whether arithmetic, geometric, or neither), since $S_n = a_1 + a_2 + \dots + a_{n-1} + a_n = S_{n-1} + a_n$, you can immediately extract the $n$-th term by subtracting consecutive partial sums:

$$a_1 = S_1, \qquad a_n = S_n - S_{n-1} \quad (n \ge 2)$$

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Summary Reference Table

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Worked Examples

Worked Example 1: Telescoping Logarithms & Fractional Base Inequalities

Problem:

1. Evaluate $P = \log_2 3 \cdot \log_3 5 \cdot \log_5 8 + 9^{\log_3 5}$.
2. Find the number of integers $x$ satisfying $\log_{1/2}(x - 3) \ge -2$.

Step-by-Step Solution:

1. In the product $\log_2 3 \cdot \log_3 5 \cdot \log_5 8$, the intermediate bases and arguments cancel via the change-of-base chain rule:

$$\log_2 3 \cdot \log_3 5 \cdot \log_5 8 = \log_2 8 = \log_2(2^3) = 3$$

For the second term, apply the exponent-swap identity $a^{\log_b c} = c^{\log_b a}$ (or rewrite $9 = 3^2$):

$$9^{\log_3 5} = 5^{\log_3 9} = 5^2 = 25$$

Therefore, $P = 3 + 25 = 28$. 2. For the inequality $\log_{1/2}(x - 3) \ge -2$:

• Step A (Domain): The argument must be strictly positive: $x - 3 > 0 \implies x > 3$.
• Step B (Unwrap base $a = 1/2 < 1$): Since $0 < 1/2 < 1$, removing $\log_{1/2}$ reverses the inequality sign:

$$x - 3 \le \left(\frac{1}{2}\right)^{-2} = 4 \implies x \le 7$$

• Combining $x > 3$ and $x \le 7$ gives $x \in (3, 7]$, which contains the integers $\{4, 5, 6, 7\}$ — a total of $4$ integers.

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Worked Example 2: Complex Powers & Sequence Partial Sums

Problem:

1. Simplify $z = \left(\frac{1 + i}{1 - i}\right)^{2025} + (1 - i)^4$.
2. In a sequence $(a_n)$, the sum of the first $n$ terms is $S_n = 2n^2 + 3n$. Find $a_8$.

Step-by-Step Solution:

1. Use the high-speed complex identities:
   • Since $\frac{1 + i}{1 - i} = i$, the first term is $i^{2025}$. Because $2025 = 4(506) + 1$, we have $i^{2025} = i^1 = i$.
   • Since $(1 - i)^2 = -2i$, we have $(1 - i)^4 = [(1 - i)^2]^2 = (-2i)^2 = 4i^2 = -4$.
   • Combining both terms yields $z = -4 + i$.
2. Rather than finding $a_1$ and $d$ separately, use the partial sum difference identity $a_n = S_n - S_{n-1}$ directly at $n = 8$:

$$S_8 = 2(8)^2 + 3(8) = 128 + 24 = 152$$ $$S_7 = 2(7)^2 + 3(7) = 98 + 21 = 119$$ $$a_8 = S_8 - S_7 = 152 - 119 = 33$$

Constant and Zero-Ratio Sequences Need Separate Checks

The constant sequence 3, 3, 3 is arithmetic with difference zero and geometric with ratio one. Its first $n$ terms sum to $3n$; a geometric sum formula written with denominator $r-1$ cannot be used at $r=1$. Recognizing the constant case avoids turning a simple sum into an undefined fraction. A classification question must therefore allow overlap unless its wording excludes one of these cases.

The sequence 5, 0, 0, 0 is geometric under the recurrence $a_{n+1}=r a_n$ with $r=0$. However, dividing the third term by the second gives $0/0$, so that ratio calculation cannot determine $r$. Use the first transition, where the preceding term is nonzero, and state the first term separately. Every later term is zero and every nonempty partial sum is 5. Likewise, an all-zero sequence satisfies this recurrence for any ratio; its terms alone do not identify a unique ratio. These examples show why recurrence definitions and domain checks are more reliable than automatically dividing consecutive terms.

Review the Method

Whenever a sequence's partial sum formula S_n is given, individual terms are recovered via a_1 = S_1 and a_n = S_n - S_{n-1} for n >= 2.