Math Problem Statement

Find the value of k in each of the following: (i) (2/3)^3 × (2/3)^6 = (4/9)^(2k-3), (ii) (-4/5)^2 × (4/5)^5 = (4/5)^(6k+1).

Solution

Let's solve each part of question 26 by finding the value of kk that satisfies the given equations.

Part (i)

The equation given is: (23)3×(23)6=(49)2k3\left( \frac{2}{3} \right)^3 \times \left( \frac{2}{3} \right)^6 = \left( \frac{4}{9} \right)^{2k-3}

Step-by-Step Solution

  1. Combine the terms on the left side: (23)3×(23)6=(23)3+6=(23)9\left( \frac{2}{3} \right)^3 \times \left( \frac{2}{3} \right)^6 = \left( \frac{2}{3} \right)^{3+6} = \left( \frac{2}{3} \right)^9

  2. Rewrite 49\frac{4}{9} as (23)2\left( \frac{2}{3} \right)^2: (49)2k3=((23)2)2k3=(23)2(2k3)=(23)4k6\left( \frac{4}{9} \right)^{2k-3} = \left( \left( \frac{2}{3} \right)^2 \right)^{2k-3} = \left( \frac{2}{3} \right)^{2(2k-3)} = \left( \frac{2}{3} \right)^{4k-6}

  3. Equate the exponents since the bases are the same: 9=4k69 = 4k - 6

  4. Solve for kk: 4k=15k=154=3.754k = 15 \Rightarrow k = \frac{15}{4} = 3.75

So, the value of kk for part (i) is: k=3.75k = 3.75


Part (ii)

The equation given is: (45)2×(45)5=(45)6k+1\left( -\frac{4}{5} \right)^2 \times \left( \frac{4}{5} \right)^5 = \left( \frac{4}{5} \right)^{6k+1}

Step-by-Step Solution

  1. Simplify the left side: (45)2=(45)2\left( -\frac{4}{5} \right)^2 = \left( \frac{4}{5} \right)^2 So, (45)2×(45)5=(45)2+5=(45)7\left( \frac{4}{5} \right)^2 \times \left( \frac{4}{5} \right)^5 = \left( \frac{4}{5} \right)^{2+5} = \left( \frac{4}{5} \right)^7

  2. Equate the exponents since the bases are the same: 7=6k+17 = 6k + 1

  3. Solve for kk: 6k=6k=16k = 6 \Rightarrow k = 1

So, the value of kk for part (ii) is: k=1k = 1


Summary

  • For part (i), k=3.75k = 3.75
  • For part (ii), k=1k = 1

Would you like a more detailed explanation of any step or have questions about these solutions?


Related Questions

  1. How would you solve equations with variables in both the base and exponent?
  2. What properties of exponents are used when solving exponential equations like these?
  3. How can we simplify expressions involving fractional exponents?
  4. Why does the base need to be the same when equating exponents?
  5. What are some real-world applications of exponential equations?

Tip

When simplifying exponent expressions, always look for common bases, as it allows you to equate the exponents and solve for unknowns.

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Math Problem Analysis

Mathematical Concepts

Exponents
Algebraic Equations
Exponent Laws

Formulas

(a^m) × (a^n) = a^(m+n)
(a^m)^n = a^(m*n)

Theorems

Laws of Exponents

Suitable Grade Level

Grades 9-10