9.1 Exponential Equations (Üslü Sayılar), Nested Radicals & Rationalizing Denominators

Exponential Equations (Üslü Sayılar) & Radical Expressions (Köklü Sayılar)

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Core Laws of Exponents (Üslü Sayılar) & The Parentheses Sign Trap

For real bases $a, b \ne 0$ and integers $m, n$, all exponential manipulations reduce to five foundational identities:

$$a^m \cdot a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{m \cdot n}$$ $$(a \cdot b)^n = a^n \cdot b^n, \qquad a^{-n} = \frac{1}{a^n}, \qquad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$$

1. The Parentheses Sign Trap (Parantez ve İşaret Tuzağı)

An even exponent ($2k$) turns a negative sign into a positive sign if and only if the exponent sits outside parentheses that enclose the negative sign:

$$(-a)^{2k} = +a^{2k} \qquad \text{vs.} \qquad -a^{2k} = -(a^{2k}) \quad \text{and} \quad (-a^{2k}) = -a^{2k}$$

When simplifying chained powers such as $(-x^2)^3 \cdot (-x^3)^2 \cdot (-x)^{-4}$ for $x \ne 0$, determine the overall sign first before combining exponents:

• In $(-x^2)^3$, the base inside the outer cube is negative ($-x^2$), and an odd outer power keeps it negative: $-x^6$.
• In $(-x^3)^2$, the outer power $2$ is even and encloses the minus sign, making the term positive: $+x^6$.
• In $(-x)^{-4}$, the outer power $-4$ is an even integer enclosing the minus sign, making the term positive: $+x^{-4}$.
• Multiplying the three terms gives $(-x^6)(+x^6)(+x^{-4}) = -x^{6 + 6 - 4} = -x^8$.

2. Factoring Exponential Sums & Differences (Ortak Çarpan Parantezine Alma)

There is no direct formula for adding $a^m + a^n$. Whenever powers of the same base are added or subtracted in a numerator and denominator, split compound exponents via $a^{x+k} = a^x \cdot a^k$ and factor out the smallest exponent:

$$2^{x+3} + 2^{x+1} - 2^x = 2^x(2^3 + 2^1 - 1) = 2^x(8 + 2 - 1) = 9 \cdot 2^x$$

With negative or fractional exponents, remember that $-x - 2$ is smaller than $-x$: factoring $3^{-x-2}$ from $3^{-x} + 3^{-x-1} + 3^{-x-2}$ gives $3^{-x-2}(3^2 + 3^1 + 1) = 13 \cdot 3^{-x-2}$.

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Exponential Equations (Üslü Denklemler) & The 3-Case Power-One Rule

1. Base Equality vs. Exponent Equality

• Same Base ($a^{f(x)} = a^{g(x)}$): If $a \in \mathbb{R}^+ \setminus \{1\}$ (so $a \notin \{0, 1, -1\}$), you can equate the exponents directly: $f(x) = g(x)$.
• Same Odd Exponent ($[f(x)]^{2n+1} = [g(x)]^{2n+1}$): Equate the bases directly: $f(x) = g(x)$.
• Same Even Exponent ($[f(x)]^{2n} = [g(x)]^{2n}$ for $n \in \mathbb{Z}^+$): Taking an even root introduces absolute values ($\lvert f(x) \rvert = \lvert g(x) \rvert$), so you must split into two branches:

$$f(x) = g(x) \qquad \text{or} \qquad f(x) = -g(x)$$

2. The 3-Case Power-One Equation: $[b(x)]^{f(x)} = 1$

1. Case 1 — Zero Exponent ($a^0 = 1$): Set $f(x) = 0$, and verify that $b(x) \ne 0$ at each candidate root (because $0^0$ is undefined!).
2. Case 2 — Unit Base ($1^{\text{any}} = 1$): Set $b(x) = 1$. Every real root of $b(x) = 1$ is automatically valid.
3. Case 3 — Negative Unit Base with Even Exponent ($(-1)^{2k} = 1$): Set $b(x) = -1$, then substitute each candidate root into $f(x)$ and keep it only if $f(x)$ is an even integer (or a reduced fraction $\frac{p}{q}$ with even numerator $p$ and odd denominator $q$).

3. Proportion Law for Coprime Bases

When two equations equate powers of two coprime positive bases $a,b$ different from 1 (such as $2$ and $3$), line up the same bases on the same sides:

$$\begin{cases} a^x = b^y \\ a^z = b^t \end{cases} \implies x\cdot t=y\cdot z \quad (\text{ratios }x/z=y/t\text{ require }z,t\ne0)$$

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Radical Expressions (Köklü Sayılar), Nested Roots & Conjugates

1. Real Domain & Even-Root Cancellation (Mutlak Değer Kuralı)

In the set of real numbers $\mathbb{R}$, an even-indexed root $\sqrt[2n]{x}$ is defined if and only if $x \ge 0$, and its value is always non-negative ($\sqrt[2n]{x} \ge 0$). If a single expression contains both $\sqrt{x - k}$ and $\sqrt{k - x}$ in $\mathbb{R}$, then $x - k \ge 0$ and $k - x \ge 0$ simultaneously force $x = k$!

When canceling a root index against an inner exponent, even indices require absolute value bars, while odd indices preserve the original sign:

$$\sqrt[2n]{x^{2n}} = \lvert x \rvert = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} \qquad \text{whereas} \qquad \sqrt[2n+1]{x^{2n+1}} = x$$

2. Fractional Exponents & Nested Root Compression

For nonnegative bases, a radical can be rewritten as a fractional exponent via $\sqrt[n]{a^m} = a^{m/n}$. For nonnegative factors and integer root indices at least 2, when roots are nested purely by multiplication, multiply the root indices and push outer factors into the innermost root by raising them to each inner index they cross:

$$\sqrt[m]{a\sqrt[n]{b\sqrt[k]{c}}} = \sqrt[m \cdot n \cdot k]{a^{n \cdot k} \cdot b^k \cdot c}$$

3. The Double-Radical Identity ($\sqrt{a \pm 2\sqrt{b}}$ Kuralı)

When a square root has the exact form $\sqrt{a \pm 2\sqrt{b}}$ (note the mandatory coefficient $2$ directly in front of the inner square root), find two positive numbers $x_1 > x_2 > 0$ whose sum is $a$ and whose product is $b$:

$$\underbrace{(x_1 + x_2)}_{a} \quad \text{and} \quad \underbrace{(x_1 \cdot x_2)}_{b} \implies \sqrt{(x_1 + x_2) \pm 2\sqrt{x_1 x_2}} = \sqrt{x_1} \pm \sqrt{x_2} \quad (x_1 > x_2 > 0)$$

Always write the larger root $\sqrt{x_1}$ first so that $\sqrt{x_1} - \sqrt{x_2} > 0$.

• If the inner coefficient is an even multiple larger than $2$ (e.g., $\sqrt{11 - 4\sqrt{6}}$): factor out $2$ and push the remaining factor inside the square root: $4\sqrt{6} = 2 \cdot 2\sqrt{6} = 2\sqrt{24}$. Since $8 + 3 = 11$ and $8 \times 3 = 24$, $\sqrt{11 - 2\sqrt{24}} = \sqrt{8} - \sqrt{3} = 2\sqrt{2} - \sqrt{3}$.
• If the inner coefficient $2$ is missing (e.g., $\sqrt{4 + \sqrt{7}}$): multiply and divide the radicand by $2$ to create $2\sqrt{7}$: $\sqrt{\frac{8 + 2\sqrt{7}}{2}} = \frac{\sqrt{7} + \sqrt{1}}{\sqrt{2}} = \frac{\sqrt{14} + \sqrt{2}}{2}$.

4. Infinite Nested Radicals (Sonsuz Köklü İfadeler)

For convergent infinite periodic radicals with $a > 0$:

• Infinite Product: $\sqrt[n]{a\sqrt[n]{a\sqrt[n]{a\dots}}} = \sqrt[n-1]{a}$ (for square roots $n = 2$, $\sqrt{a\sqrt{a\sqrt{a\dots}}} = a$).
• Infinite Quotient: $\sqrt[n]{a \div \sqrt[n]{a \div \sqrt[n]{a \div \dots}}} = \sqrt[n+1]{a}$ (for $n = 2$, $\sqrt{a \div \sqrt{a \div \dots}} = \sqrt[3]{a}$).
• Consecutive-Factor Infinite Sum / Difference ($A = k(k+1)$ with $k > 0$):

$$\sqrt{k(k+1) + \sqrt{k(k+1) + \dots}} = k + 1, \qquad \sqrt{k(k+1) - \sqrt{k(k+1) - \dots}} = k$$

5. Rationalizing Denominators (Paydayı Rasyonel Yapma) & Telescoping Sums

To eliminate square roots from a denominator $\sqrt{a} \pm \sqrt{b}$, multiply numerator and denominator by its conjugate (eşlenik) $\sqrt{a} \mp \sqrt{b}$, using $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$:

$$\frac{c}{\sqrt{a} \pm \sqrt{b}} = \frac{c(\sqrt{a} \mp \sqrt{b})}{a - b}$$

When $a = k + 1$ and $b = k$ are consecutive integers, $a - b = 1$ and $\frac{1}{\sqrt{k+1} + \sqrt{k}} = \sqrt{k+1} - \sqrt{k}$. Summing from $k = 1$ to $n$ produces a telescoping cancellation where every intermediate root cancels:

$$\sum_{k=1}^{n} \frac{1}{\sqrt{k+1} + \sqrt{k}} = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \dots + (\sqrt{n+1} - \sqrt{n}) = \sqrt{n+1} - 1$$

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Summary Table of Exponent & Radical Shortcuts

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Worked Examples

Worked Example 1: The 3-Case Power-One Equation

Problem: Find the sum of all distinct real roots of the equation $(x^2 - 5x + 5)^{x^2 - 9} = 1$.

Solution: We test all three cases of $[b(x)]^{f(x)} = 1$ where $b(x) = x^2 - 5x + 5$ and $f(x) = x^2 - 9$:

1. Case 1 (Zero Exponent): $f(x) = x^2 - 9 = 0 \implies x = 3$ or $x = -3$.
   • Check $b(3) = 9 - 15 + 5 = -1 \ne 0$ (valid) and $b(-3) = 9 + 15 + 5 = 29 \ne 0$ (valid).
2. Case 2 (Unit Base): $b(x) = x^2 - 5x + 5 = 1 \implies x^2 - 5x + 4 = (x - 1)(x - 4) = 0 \implies x = 1$ or $x = 4$ (both valid).
3. Case 3 (Negative Unit Base with Even Exponent): $b(x) = x^2 - 5x + 5 = -1 \implies x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \implies x = 2$ or $x = 3$.
   • At $x = 2$, the exponent is $f(2) = 2^2 - 9 = -5$, which is odd, giving $(-1)^{-5} = -1 \ne 1$, so $x = 2$ is rejected.
   • At $x = 3$, the exponent is $f(3) = 0$ (even, already counted in Case 1).

The distinct real solution set is $\{-3, 1, 3, 4\}$, and the sum of the roots is $(-3) + 1 + 3 + 4 = \mathbf{5}$.

Worked Example 2: Double-Radical Simplification & Conjugate Rationalization

Problem: Evaluate $\frac{6}{\sqrt{7 - 4\sqrt{3}}} - \frac{2}{\sqrt{3} + 1}$.

Solution:

1. Rewrite the inner coefficient $4\sqrt{3}$ as $2 \cdot 2\sqrt{3} = 2\sqrt{12}$ so the double-radical identity applies:

$$\sqrt{7 - 4\sqrt{3}} = \sqrt{7 - 2\sqrt{12}} = \sqrt{(4 + 3) - 2\sqrt{4 \cdot 3}} = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3}$$

1. Rationalize both denominators using their conjugates $(2 + \sqrt{3})$ and $(\sqrt{3} - 1)$:

$$\frac{6}{2 - \sqrt{3}} - \frac{2}{\sqrt{3} + 1} = \frac{6(2 + \sqrt{3})}{4 - 3} - \frac{2(\sqrt{3} - 1)}{3 - 1} = 6(2 + \sqrt{3}) - (\sqrt{3} - 1) = 12 + 6\sqrt{3} - \sqrt{3} + 1 = \mathbf{13 + 5\sqrt{3}}$$

The exponent-proportion cross-product identity follows from taking logarithms. Ratios of exponents require nonzero denominators. Conjugate rationalization requires the original denominator and $a-b$ to be nonzero; if $a=b>0$, simplify a sum denominator directly. Infinite-root identities assume convergence and valid real radicands.

Review the Method

Under the Double-Radical Identity, if $x_1 + x_2 = a$ and $x_1 \cdot x_2 = b$ with $x_1 > x_2 > 0$, then $\sqrt{a \pm 2\sqrt{b}} = \sqrt{x_1} \pm \sqrt{x_2}$.