If $\frac{1}{64}=4^{2s-1} \cdot 16^{2s+2}$, What Is The Value Of $s$?A. \[$-1\$\]B. \[$0\$\]C. \[$1\$\]D. No Solution

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Introduction

In this article, we will explore a mathematical equation involving exponents and solve for the value of a variable, s. The equation given is 164=42s−1⋅162s+2\frac{1}{64}=4^{2s-1} \cdot 16^{2s+2}. Our goal is to isolate the variable s and determine its value.

Understanding the Equation

Before we begin solving the equation, let's break it down and understand its components. The equation involves two bases: 4 and 16. We can rewrite 16 as 424^2, which will help us simplify the equation.

Rewriting the Equation

We can rewrite the equation as follows:

164=42s−1⋅(42)2s+2\frac{1}{64}=4^{2s-1} \cdot (4^2)^{2s+2}

Using the property of exponents that (ab)c=abc(a^b)^c = a^{bc}, we can simplify the right-hand side of the equation:

164=42s−1⋅44s+4\frac{1}{64}=4^{2s-1} \cdot 4^{4s+4}

Now, we can combine the two terms on the right-hand side using the property of exponents that abâ‹…ac=ab+ca^b \cdot a^c = a^{b+c}:

164=42s−1+4s+4\frac{1}{64}=4^{2s-1+4s+4}

Simplifying the exponent, we get:

164=46s+3\frac{1}{64}=4^{6s+3}

Solving for s

Now that we have simplified the equation, we can set the exponents equal to each other:

−6=6s+3-6=6s+3

Subtracting 3 from both sides, we get:

−9=6s-9=6s

Dividing both sides by 6, we get:

−32=s-\frac{3}{2}=s

Conclusion

In this article, we solved a mathematical equation involving exponents and determined the value of the variable s. The equation was 164=42s−1⋅162s+2\frac{1}{64}=4^{2s-1} \cdot 16^{2s+2}, and we found that the value of s is −32-\frac{3}{2}. This solution is unique and satisfies the given equation.

Answer

The correct answer is:

A. {-\frac{3}{2}$}$

Discussion

This problem requires a strong understanding of exponents and algebraic manipulation. The key to solving this problem is to simplify the equation by combining the terms on the right-hand side and then setting the exponents equal to each other. The solution is unique and can be verified by plugging it back into the original equation.

Additional Tips

  • When solving equations involving exponents, it's essential to simplify the equation by combining the terms and setting the exponents equal to each other.
  • Make sure to use the properties of exponents correctly, such as (ab)c=abc(a^b)^c = a^{bc} and abâ‹…ac=ab+ca^b \cdot a^c = a^{b+c}.
  • Verify the solution by plugging it back into the original equation to ensure that it satisfies the equation.
    Q&A: Solving for the Value of s in a Given Equation =====================================================

Introduction

In our previous article, we solved a mathematical equation involving exponents and determined the value of the variable s. The equation was 164=42s−1⋅162s+2\frac{1}{64}=4^{2s-1} \cdot 16^{2s+2}, and we found that the value of s is −32-\frac{3}{2}. In this article, we will answer some frequently asked questions related to this problem.

Q: What is the main concept behind solving this equation?

A: The main concept behind solving this equation is to simplify the equation by combining the terms on the right-hand side and then setting the exponents equal to each other. This requires a strong understanding of exponents and algebraic manipulation.

Q: How do I simplify the equation?

A: To simplify the equation, you can combine the two terms on the right-hand side using the property of exponents that abâ‹…ac=ab+ca^b \cdot a^c = a^{b+c}. This will help you to rewrite the equation in a simpler form.

Q: What is the property of exponents that I should use to simplify the equation?

A: The property of exponents that you should use to simplify the equation is (ab)c=abc(a^b)^c = a^{bc}. This will help you to rewrite the equation in a simpler form.

Q: How do I set the exponents equal to each other?

A: To set the exponents equal to each other, you can equate the exponents on both sides of the equation. This will give you an equation in terms of the variable s.

Q: What is the final step in solving the equation?

A: The final step in solving the equation is to solve for the variable s. This involves isolating the variable s and determining its value.

Q: How do I verify the solution?

A: To verify the solution, you can plug it back into the original equation to ensure that it satisfies the equation.

Q: What are some common mistakes to avoid when solving this type of equation?

A: Some common mistakes to avoid when solving this type of equation include:

  • Not simplifying the equation properly
  • Not using the properties of exponents correctly
  • Not setting the exponents equal to each other
  • Not solving for the variable s correctly

Q: Can you provide an example of a similar equation?

A: Yes, here is an example of a similar equation:

181=32s−2⋅92s+1\frac{1}{81}=3^{2s-2} \cdot 9^{2s+1}

This equation can be solved using the same steps as the original equation.

Conclusion

In this article, we answered some frequently asked questions related to solving the equation 164=42s−1⋅162s+2\frac{1}{64}=4^{2s-1} \cdot 16^{2s+2}. We provided tips and examples to help you understand the concept and solve similar equations.

Additional Tips

  • Make sure to simplify the equation properly before setting the exponents equal to each other.
  • Use the properties of exponents correctly to simplify the equation.
  • Verify the solution by plugging it back into the original equation.
  • Practice solving similar equations to improve your skills.

Resources

  • Khan Academy: Exponents and Exponential Functions
  • Mathway: Exponents and Exponential Functions
  • Wolfram Alpha: Exponents and Exponential Functions