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Chapter 13: Problem 39

Solve each exponential equation. $$32^{3 c}=8^{c+4}$$

Short Answer

Expert verified
To solve the exponential equation \(32^{3c} = 8^{c+4}\), rewrite the bases as powers of a common base: \((2^5)^{3c} = (2^3)^{c+4}\). Apply the logarithmic properties to simplify the equation: \(2^{15c} =2^{3c+12}\). Since the base numbers are the same, equate the exponents to solve for the variable \(c\): \(15c = 3c+12\). Solving for \(c\) gives \(c=1\).

Step by step solution

01

Rewrite the bases as powers of a common base

Let's rewrite the given equation using a common base \(2\). From this, we know that \(32 = 2^5\) and \(8 = 2^3\), so we can replace the base numbers in the equation with these powers of \(2\). \( (2^5)^{3c} = (2^3)^{c+4} \)
02

Apply the logarithmic property to simplify the equation

Now, let's apply the logarithmic properties to simplify the equation. By using the Power Rule (which states that \( (a^{m})^n = a^{m \times n} \)), we can rewrite the equation as: \( 2^{5 \times 3c} = 2^{3 \times (c+4)} \) Which simplifies to: \( 2^{15c} =2^{3c+12}\)
03

Solve for the variable \(c\)

Since the base numbers are the same on both sides of the equation (base \(2\)), we can equate the exponents to solve for the variable \(c\). This gives us: \( 15c = 3c+12 \) Now, let's solve for \(c\). Subtract \(3c\) from both sides of the equation: \( 12c = 12 \) Now divide both sides by \(12\): \( c = \frac{12}{12} \) This gives us the solution for the variable \(c\): \( c = 1 \) So, the solution to the exponential equation \(32^{3c} = 8^{c+4}\) is \(c=1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Logarithmic Properties
Logarithmic properties are key concepts in mathematics that let us manipulate exponents and solve equations easier. Using logarithms can help turn complex multiplication and division into simpler addition and subtraction. Important properties include:
  • Product Property: \( \log_b(M \times N) = \log_b M + \log_b N \)
  • Quotient Property: \( \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N \)
  • Power Property: \( \log_b(M^n) = n \cdot \log_b M \)
These properties are helpful for working with exponential equations by allowing transformation of large powers and products into more manageable forms. In our problem, while we didn't explicitly use logarithms, understanding these properties underpins why substituting logarithm-based simplifications happens in exponential equations.
Exploring the Power Rule
The Power Rule is a fundamental concept when working with exponents and powers. It states that when you raise an exponent to another power, you multiply the exponents. Mathematically, it is expressed as:
  • \( (a^m)^n = a^{m \times n} \)
This rule is essential in simplifying expressions where bases are raised to additional power. In the problem of solving the equation \((2^5)^{3c} = (2^3)^{c+4}\), we applied the power rule to both sides. This transformed it into \(2^{15c} = 2^{3c + 12}\), allowing us to easily compare and solve the exponents directly. By understanding how to use the power rule, complex exponent expressions can be broken down and solved.
Identifying the Common Base
Choosing the right common base is crucial when solving exponential equations. A common base means that two different numbers are expressed as powers of the same number. In this equation, we transformed \(32\) and \(8\) into powers of \(2\):
  • \(32 = 2^5\)
  • \(8 = 2^3\)
By reducing the problem to a common base, we make the equations easier to handle by directly focusing on the exponents since the base is uniform. Solving becomes simpler as the focus shifts from the base to resolving the exponents of each side of the equation. This method is especially useful in various mathematical and practical applications, from algebra to physics.
Solving for the Variable
Once exponential expressions have the same base, solving for the variable relies on simplifying the equation to find what value of the variable satisfies it. In our equation after rewriting, since both sides had the same base \(2\), the problem reduced to comparing exponents:
  • \(15c = 3c + 12\)
By isolating the variable, you subtract \(3c\) from both sides to gain:
  • \(12c = 12\)
Further solving for \(c\) by dividing both sides by 12 results in \(c = 1\). This demonstrates how equating exponents upon having a common base opens a straightforward path to solving for unknown variables, making initially complex-looking problems much more manageable.

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Most popular questions from this chapter

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