Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Solve the equation.
![**Problem Statement:**
Solve the equation:
\[ 64^{4x} = 16^{x+3} \]
**Solution Set**:
The given equation involves exponential expressions with the same base (powers of 2). The base 64 can be rewritten as \( 2^6 \) and 16 as \( 2^4 \).
Rewriting the equation in terms of the base 2:
\[ (2^6)^{4x} = (2^4)^{x+3} \]
Simplify the exponents:
\[ 2^{24x} = 2^{4(x+3)} \]
Which simplifies to:
\[ 24x = 4x + 12 \]
By solving the linear equation:
1. Subtract \( 4x \) from both sides:
\[ 20x = 12 \]
2. Divide both sides by 20:
\[ x = \frac{12}{20} \]
3. Simplify the fraction:
\[ x = \frac{3}{5} \]
Thus, the solution set is:
\[ x = \frac{3}{5} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fa757ec-9ef0-41e2-a2d2-ee98e8402c6d%2F06164091-3f0f-49e8-8f71-a7f29ba3fbb5%2Fb0oyob_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Solve the equation:
\[ 64^{4x} = 16^{x+3} \]
**Solution Set**:
The given equation involves exponential expressions with the same base (powers of 2). The base 64 can be rewritten as \( 2^6 \) and 16 as \( 2^4 \).
Rewriting the equation in terms of the base 2:
\[ (2^6)^{4x} = (2^4)^{x+3} \]
Simplify the exponents:
\[ 2^{24x} = 2^{4(x+3)} \]
Which simplifies to:
\[ 24x = 4x + 12 \]
By solving the linear equation:
1. Subtract \( 4x \) from both sides:
\[ 20x = 12 \]
2. Divide both sides by 20:
\[ x = \frac{12}{20} \]
3. Simplify the fraction:
\[ x = \frac{3}{5} \]
Thus, the solution set is:
\[ x = \frac{3}{5} \]
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