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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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MathematicsAlgebra
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Given that 50.6=x,50.2=y5^{0.6} = x, 5^{0.2} = y50.6=x,50.2=y and xz=y2x^{z} = y^{2}xz=y2, then the value of z is:

A

2/3

B

3/4

C

4/5

D

5/6

Correct Answer

📐 MA • Math Exponent Power RuleMathematicsAlgebra
Option A

2/3

Quick Summary:

Express both xxx and yyy as powers of 555, then equate the exponents directly by ignoring the base 5.

📐MAMath SolutionExponent Power Rule
📋 Given

Given x=50.6x = 5^{0.6}x=50.6 and y=50.2y = 5^{0.2}y=50.2, with the equation xz=y2x^{z} = y^{2}xz=y2.

🔢 Formula Used

(am)n=am×n(a^m)^n = a^{m \times n}(am)n=am×n

⚡ Exam Hall Shortcut / Speed Trick

Express both xxx and yyy as powers of 555, then equate the exponents directly by ignoring the base 5.

⚠️ Common Student Trap / Pitfall

Confusing 0.60.60.6 and 0.20.20.2 with fractions or failing to multiply the power of yyy by 222, which leads to z=0.4z=0.4z=0.4 instead of z=0.666...z=0.666...z=0.666....

📊 Diagram / Illustration
Algebraic Exponent Solution 1 Given Data & Substitution x = 5⁰.6, y = 5⁰.2 | Equation: (5⁰.6)ᶻ = (5⁰.2)² 2 Apply Power Rule: (aᵐ)ⁿ = a^(m×n) 5^(0.6 × z) = 5^(0.2 × 2) ⇒ 5⁰.6z = 5⁰.4 3 Equate Exponents 0.6z = 0.4 ⇒ z = 0.4 / 0.6 = 4 / 6 = 2 / 3 4 Final Result z = 2/3 Final Answer: z = 2/3 (Option A)
🔢 Step-by-Step Solution
1

Convert x and y to powers of 5

Substitute the given values of xxx and yyy into the equation xz=y2x^{z} = y^{2}xz=y2.

(50.6)z=(50.2)2(5^{0.6})^{z} = (5^{0.2})^{2}(50.6)z=(50.2)2

2

Apply power of power rule

Using (am)n=am×n(a^m)^n = a^{m \times n}(am)n=am×n, expand both sides.

50.6z=50.45^{0.6z} = 5^{0.4}50.6z=50.4

3

Equate the exponents

Since the bases are equal, the exponents must be equal: 0.6z=0.40.6z = 0.40.6z=0.4.

z=0.40.6=46=23z = \frac{0.4}{0.6} = \frac{4}{6} = \frac{2}{3}z=0.60.4​=64​=32​

✅

A is correct because solving the equation 50.6z=50.45^{0.6z} = 5^{0.4}50.6z=50.4 yields z=2/3z = 2/3z=2/3.

Core Concepts Used
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Laws of Exponents Base Equality Principle Linear Equations in one variable
💡 EXAM TIP

This concept is fundamental to solving logarithmic equations and base-change problems in quantitative aptitude exams.

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