Exponent & Root: Level 3

Solution (D): Convert all terms to base 5. \(125 = 5^3\) and \(25 = 5^2\). \( \frac{(5^3)^{x+2}}{(5^2)^{2x-1}} = 5^1 \) \( \frac{5^{3x+6}}{5^{4x-2}} = 5^1 \) \( 5^{(3x+6) – (4x-2)} = 5^1 \) \( 5^{3x+6 – 4x+2} = 5^1 \) \( 5^{-x+8} = 5^1 \) Equate exponents: \(-x + 8 = 1 \implies 7 = x\).

Solution (B): Factor out the smallest term, \(4^x\). \( 4^x (4^2 – 1) = 240 \) \( 4^x (16 – 1) = 240 \) \( 4^x (15) = 240 \) \( 4^x = \frac{240}{15} = 16 \) Since \( 16 = 4^2 \), \(x = 2\).

Solution (D): 1. Analyze the given: \(9^x = (3^2)^x = 3^{2x} = 100\). 2. Simplify the expression: \( (3^{x-1})^2 = 3^{2(x-1)} = 3^{2x-2} \). 3. Use exponent rules: \( 3^{2x-2} = 3^{2x} \times 3^{-2} = \frac{3^{2x}}{3^2} \). 4. Substitute the given value: \( \frac{100}{3^2} = \frac{100}{9} \).

Solution (B): 1. Break all bases into primes: \( 20^x = (4 \times 5)^x = (2^2 \times 5)^x = 2^{2x} \times 5^x \). 2. \( 4^{y+1} = (2^2)^{y+1} = 2^{2y+2} \). 3. The full equation is: \( (5^{2x-1})(2^{2y+2}) = 2^{2x} \times 5^x \). 4. Compare exponents for base 5: \( 2x – 1 = x \implies x = 1 \). 5. Compare exponents for base 2: \( 2y + 2 = 2x \). 6. Substitute \(x=1\): \( 2y + 2 = 2(1) \implies 2y = 0 \implies y = 0 \). 7. \( x + y = 1 + 0 = 1 \).

Solution (B): 1. Convert all terms to base 2: \( 4^{31} = (2^2)^{31} = 2^{62} \). 2. The equation is \( 2^{62} + 2^{60} = 2^x \). 3. Factor out the smaller term, \( 2^{60} \): \( 2^{60}(2^2 + 1) = 2^x \) \( 2^{60}(4 + 1) = 2^x \) \( 5 \times 2^{60} = 2^x \) 4. This is an approximation. We know \( 4 = 2^2 \) and \( 8 = 2^3 \). 5. \( 5 \times 2^{60} \) is between \( 4 \times 2^{60} \) (which is \( 2^2 \times 2^{60} = 2^{62} \)) and \( 8 \times 2^{60} \) (which is \( 2^3 \times 2^{60} = 2^{63} \)). 6. Since 5 is closer to 4 than it is to 8, the value is best approximated by \( 2^{62} \). So, \(x \approx 62\).

Solution (D): This is a logic puzzle from image `f969ab`. \( \frac{15^{x+x+1}}{4^y} = 15^y \implies \frac{15^{2x+1}}{4^y} = 15^y \) \( 15^{2x+1} = 15^y \times 4^y \). \( 15^{2x+1} = (15 \times 4)^y = 60^y \). This cannot be true unless both sides are 1. Let’s re-read the image: \( \frac{15^x \times 15^{x+1}}{4^y} = 15^y \). My algebra `15^(2x+1) = 15^y * 4^y` is correct. This equation `15^(2x+1) = 60^y` cannot be solved. Let’s re-read the image *very* carefully. ` (15^x * 15^{x+1}) / 4^y = 15^y ` This is what I have. Wait, the image `f969ab` Q5 is `(15^x * 15^{x+1}) / 4^y = 15^y`. Wait, is it `4y` or `4^y`? It looks like `4y`. Let’s try that: `(15^{2x+1}) / (4y) = 15^y`. This is also not solvable. Let’s assume the question in the image is a typo and was meant to be: \( \frac{15^x + 15^{x+1}}{4y} = 15^y \). Also unsolvable. Let’s assume the image meant: `(15^x + 15^{x+1}) / (15^x) = y` `15^x(1 + 15) / 15^x = y \implies 16 = y`. Let’s try the *other* typo from the image Q10: `5^x – 5^y = (2^x)(5^{x-1})`… No that’s `(2^y)`. `5^x – 5^y = (2^y)(5^{x-1})`. Factor `5^y`… `5^y(5^{x-y} – 1) = (2^y)(5^{x-1})`. This is too complex. Let’s go back to image Q4: `2^x – 2^{x-2} = 3(2^{13})`. ` 2^{x-2}(2^2 – 1) = 3(2^{13}) ` ` 2^{x-2}(4 – 1) = 3(2^{13}) ` ` 2^{x-2}(3) = 3(2^{13}) ` ` 2^{x-2} = 2^{13} \implies x-2 = 13 \implies x = 15 `. This is a good L3 question.

Solution (D): 1. Factor out the smallest term, \(2^{x-2}\), from the left side. \( 2^{x-2} (2^2 – 1) = 3(2^{13}) \) 2. Simplify the parenthesis: \( 2^{x-2} (4 – 1) = 3(2^{13}) \) \( 2^{x-2} (3) = 3(2^{13}) \) 3. Divide both sides by 3: \( 2^{x-2} = 2^{13} \) 4. Equate the exponents: \( x – 2 = 13 \implies x = 15 \).

Solution (D): 1. Convert all terms to base 3. \(27 = 3^3\), \(9 = 3^2\), \(1 = 3^0\). 2. \( (3^3)^{4x+2} \times (3^2)^{6-x} = 3^0 \) 3. \( 3^{3(4x+2)} \times 3^{2(6-x)} = 3^0 \) 4. \( 3^{12x+6} \times 3^{12-2x} = 3^0 \) 5. Add the exponents: \( (12x+6) + (12-2x) = 0 \) 6. \( 10x + 18 = 0 \implies 10x = -18 \implies x = -1.8 \).

Solution (B): I. \(x^2 – x^2\). A number subtracted from itself is always 0. (Must be zero) II. \(x^0 – 1\). Since \(x > 0\), \(x^0 = 1\). So, \(1 – 1 = 0\). (Must be zero) III. \(x^3 – x^0 = x^3 – 1\). This is 0 only if \(x=1\). It is not *must be* zero. Therefore, only I and II must equal zero.

Solution (D): 1. Convert radicals to fractional exponents: \( \sqrt{x \cdot (x \cdot x^{1/2})^{1/2}} = 2 \) 2. \( \sqrt{x \cdot (x^{3/2})^{1/2}} = 2 \) 3. \( \sqrt{x \cdot x^{3/4}} = 2 \) 4. \( \sqrt{x^{1} \cdot x^{3/4}} = \sqrt{x^{7/4}} = 2 \) 5. \( (x^{7/4})^{1/2} = 2 \implies x^{7/8} = 2 \) 6. Raise both sides to the \(8/7\) power: \( x = 2^{8/7} \). 7. \( 2^{8/7} = \sqrt[7]{2^8} = \sqrt[7]{256} \). *Correction:* Let’s check my options. \( \sqrt[7]{16} = (2^4)^{1/7} = 2^{4/7} \). Not a match. My calculation: \(x = 2^{8/7}\). Let’s re-check. \(x \cdot x^{1/2} = x^{3/2}\). \((x^{3/2})^{1/2} = x^{3/4}\). \(x^1 \cdot x^{3/4} = x^{7/4}\). \((x^{7/4})^{1/2} = x^{7/8}\). \(x^{7/8} = 2\). \(x = 2^{8/7}\). This is correct. Let’s re-check the options. (D) \( \sqrt[7]{16} = 16^{1/7} = (2^4)^{1/7} = 2^{4/7} \). There is an error in my question. Let’s make Q8 `\sqrt{x\sqrt{x}} = 16`. `x^{3/4} = 16`. `x = 16^{4/3} = (2^4)^{4/3} = 2^{16/3}`. Ugly. Let’s use Q8 from my L3 brainstorm: `\sqrt{x\sqrt{x}} = 8`. `x^{3/4} = 8 \implies x = 8^{4/3} = (8^{1/3})^4 = 2^4 = 16`.

Solution (E): 1. Convert radicals to fractional exponents: \( \sqrt{x \cdot x^{1/2}} = 8 \). 2. Combine terms inside: \( \sqrt{x^{3/2}} = 8 \). 3. Apply the outer root: \( (x^{3/2})^{1/2} = 8 \). 4. \( x^{3/4} = 8 \). 5. To solve for \(x\), raise both sides to the \(4/3\) power: \( x = 8^{4/3} \). 6. \( 8^{4/3} = (8^{1/3})^4 = (\sqrt[3]{8})^4 = (2)^4 = 16 \).

Solution (D): This is a system of equations. 1. Eq 1 (Base 5): \( 5^x \times (5^2)^y = 5^4 \implies 5^{x+2y} = 5^4 \implies x + 2y = 4 \). 2. Eq 2 (Base 3): \( 3^x \times (3^2)^y = 3^5 \implies 3^{x+2y} = 3^5 \implies x + 2y = 5 \). *Correction:* Let’s re-read the image Q1. `(2^2x+1)(3^2y-1)`. Let’s make my Q10: `(3^x)(9^y) = 81`. 1. Eq 1 (Base 5): \(x + 2y = 4\). 2. Eq 2 (Base 3): `(3^x)(3^2)^y = 3^4 \implies x + 2y = 4`. This is the same equation. It cannot be determined. Let’s change Eq 2: \( (3^{2x})(9^y) = 243 \). 1. Eq 1 (Base 5): \(x + 2y = 4\). 2. Eq 2 (Base 3): \( 3^{2x} \times (3^2)^y = 3^5 \implies 2x + 2y = 5 \). 3. Subtract (1) from (2): \( (2x+2y) – (x+2y) = 5 – 4 \implies x = 1 \). 4. If \(x=1\), \(1 + 2y = 4 \implies 2y = 3 \implies y = 1.5\). 5. This is a valid L3 question.

Solution (B): This is a system of equations. 1. Eq 1 (Base 5): \( 5^x \times (5^2)^y = 5^4 \implies 5^{x+2y} = 5^4 \). This gives our first equation: \( x + 2y = 4 \). 2. Eq 2 (Base 3): \( 3^{2x} \times (3^2)^y = 3^5 \implies 3^{2x+2y} = 3^5 \). This gives our second equation: \( 2x + 2y = 5 \). 3. Subtract (1) from (2): \( (2x + 2y) – (x + 2y) = 5 – 4 \) \( x = 1 \).