Math Problem Statement

For the real numbers a and b, given the equation 2^a = 14^(a+b-1), find 7^a in terms of b.

Solution

The problem in the image is asking for the value of 7a7^a in terms of bb, given the equation:

2a=14a+b12^a = 14^{a + b - 1}

Let's start by breaking it down and solving step by step.

Step 1: Express 1414 as a product of primes

Since 14=2×714 = 2 \times 7, we can rewrite the equation 14a+b114^{a + b - 1} as:

2a=(2×7)a+b12^a = (2 \times 7)^{a + b - 1}

This simplifies to:

2a=2a+b1×7a+b12^a = 2^{a + b - 1} \times 7^{a + b - 1}

Step 2: Compare the powers of 2 on both sides

On the left-hand side, we have 2a2^a, and on the right-hand side, the factor involving 22 is 2a+b12^{a + b - 1}. To satisfy this equation, the powers of 2 must be equal. Therefore, we equate the powers of 2:

a=a+b1a = a + b - 1

Simplifying this equation:

0=b1b=10 = b - 1 \quad \Rightarrow \quad b = 1

Step 3: Find 7a7^a

We are asked to find 7a7^a in terms of bb. Since b=1b = 1, we substitute this back into the original equation:

2a=14a+11=14a=(2×7)a=2a×7a2^a = 14^{a + 1 - 1} = 14^a = (2 \times 7)^a = 2^a \times 7^a

Dividing both sides by 2a2^a gives:

1=7a1 = 7^a

Thus, we find:

7a=17^a = 1

Final Answer:

The value of 7a7^a is 11 when b=1b = 1.

Would you like further clarification or additional steps explained?

Here are five questions to expand on this concept:

  1. How do you solve exponential equations when variables are in the exponent?
  2. What are some common techniques for dealing with exponential equations involving different bases?
  3. Can you generalize this problem for any constant besides 14?
  4. How does logarithmic manipulation help in solving exponential equations?
  5. How can we apply this concept to real-world growth or decay problems?

Tip: When solving exponential equations, always look for ways to express numbers as powers of the same base to simplify comparisons.

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Math Problem Analysis

Mathematical Concepts

Exponential Equations
Prime Factorization
Algebra

Formulas

Prime factorization: 14 = 2 × 7
Exponent comparison
Solving for variables

Theorems

Laws of Exponents
Exponentiation and base manipulation

Suitable Grade Level

Grades 10-12