Solve The Equation:$\[ \left(\frac{1}{27}\right)^m \times (81)^{-1} = 243 \\]

by ADMIN 78 views

Introduction

In this article, we will delve into solving a complex equation involving exponents and powers. The equation is given as (127)mร—(81)โˆ’1=243\left(\frac{1}{27}\right)^m \times (81)^{-1} = 243. Our goal is to find the value of mm that satisfies this equation. We will break down the solution into manageable steps, making it easy to follow and understand.

Understanding the Equation

Before we dive into solving the equation, let's first understand its components. The equation involves two terms: (127)m\left(\frac{1}{27}\right)^m and (81)โˆ’1(81)^{-1}. We can simplify these terms to make the equation more manageable.

  • (127)m\left(\frac{1}{27}\right)^m can be rewritten as (133)m\left(\frac{1}{3^3}\right)^m, which simplifies to 133m\frac{1}{3^{3m}}.
  • (81)โˆ’1(81)^{-1} can be rewritten as (34)โˆ’1(3^4)^{-1}, which simplifies to 134\frac{1}{3^4}.

Now that we have simplified the terms, we can rewrite the equation as 133mร—134=243\frac{1}{3^{3m}} \times \frac{1}{3^4} = 243.

Simplifying the Equation

To simplify the equation further, we can combine the two fractions on the left-hand side. This gives us 133m+4=243\frac{1}{3^{3m+4}} = 243.

Next, we can rewrite 243243 as 353^5. This gives us 133m+4=35\frac{1}{3^{3m+4}} = 3^5.

Using Exponent Properties

Now that we have the equation in a simplified form, we can use exponent properties to solve for mm. Specifically, we can use the property that states amร—an=am+na^m \times a^n = a^{m+n}.

Applying this property to the equation, we get 3โˆ’(3m+4)=353^{-(3m+4)} = 3^5.

Since the bases are the same, we can equate the exponents. This gives us โˆ’(3m+4)=5-(3m+4) = 5.

Solving for m

Now that we have the equation in a simplified form, we can solve for mm. To do this, we can add 3m3m to both sides of the equation, which gives us โˆ’4=5+3m-4 = 5 + 3m.

Next, we can subtract 55 from both sides of the equation, which gives us โˆ’9=3m-9 = 3m.

Finally, we can divide both sides of the equation by 33, which gives us m=โˆ’3m = -3.

Conclusion

In this article, we solved the equation (127)mร—(81)โˆ’1=243\left(\frac{1}{27}\right)^m \times (81)^{-1} = 243 by breaking it down into manageable steps. We simplified the terms, combined the fractions, and used exponent properties to solve for mm. The final answer is m=โˆ’3m = -3.

Additional Tips and Tricks

  • When solving equations involving exponents, it's essential to simplify the terms and use exponent properties to make the equation more manageable.
  • Be careful when equating exponents, as this can lead to incorrect solutions.
  • Always check your work by plugging the solution back into the original equation.

Common Mistakes to Avoid

  • Failing to simplify the terms and use exponent properties can lead to incorrect solutions.
  • Equating exponents without considering the bases can lead to incorrect solutions.
  • Not checking the work by plugging the solution back into the original equation can lead to incorrect solutions.

Real-World Applications

Solving equations involving exponents has numerous real-world applications, including:

  • Finance: Exponents are used to calculate compound interest and investment returns.
  • Science: Exponents are used to describe the growth and decay of populations, chemical reactions, and physical systems.
  • Engineering: Exponents are used to design and optimize systems, including electrical circuits, mechanical systems, and computer networks.

Introduction

In our previous article, we solved the equation (127)mร—(81)โˆ’1=243\left(\frac{1}{27}\right)^m \times (81)^{-1} = 243 by breaking it down into manageable steps. We simplified the terms, combined the fractions, and used exponent properties to solve for mm. In this article, we will answer some common questions related to solving equations involving exponents.

Q: What are some common mistakes to avoid when solving equations involving exponents?

A: Some common mistakes to avoid when solving equations involving exponents include:

  • Failing to simplify the terms and use exponent properties can lead to incorrect solutions.
  • Equating exponents without considering the bases can lead to incorrect solutions.
  • Not checking the work by plugging the solution back into the original equation can lead to incorrect solutions.

Q: How do I simplify terms involving exponents?

A: To simplify terms involving exponents, you can use the following properties:

  • (ab)m=ambm\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}
  • (ab)m=amร—bm(ab)^m = a^m \times b^m
  • (am)n=amn(a^m)^n = a^{mn}

For example, to simplify the term (127)m\left(\frac{1}{27}\right)^m, you can rewrite it as (133)m\left(\frac{1}{3^3}\right)^m, which simplifies to 133m\frac{1}{3^{3m}}.

Q: How do I use exponent properties to solve equations?

A: To use exponent properties to solve equations, you can follow these steps:

  1. Simplify the terms and combine the fractions.
  2. Use the property that states amร—an=am+na^m \times a^n = a^{m+n} to combine the exponents.
  3. Equate the exponents and solve for the variable.

For example, to solve the equation 133m+4=35\frac{1}{3^{3m+4}} = 3^5, you can use the property that states amร—an=am+na^m \times a^n = a^{m+n} to combine the exponents. This gives you 3โˆ’(3m+4)=353^{-(3m+4)} = 3^5.

Q: What are some real-world applications of solving equations involving exponents?

A: Solving equations involving exponents has numerous real-world applications, including:

  • Finance: Exponents are used to calculate compound interest and investment returns.
  • Science: Exponents are used to describe the growth and decay of populations, chemical reactions, and physical systems.
  • Engineering: Exponents are used to design and optimize systems, including electrical circuits, mechanical systems, and computer networks.

Q: How do I check my work when solving equations involving exponents?

A: To check your work when solving equations involving exponents, you can plug the solution back into the original equation. This will help you verify that your solution is correct.

For example, to check the solution m=โˆ’3m = -3 for the equation (127)mร—(81)โˆ’1=243\left(\frac{1}{27}\right)^m \times (81)^{-1} = 243, you can plug m=โˆ’3m = -3 back into the original equation. This gives you (127)โˆ’3ร—(81)โˆ’1=243\left(\frac{1}{27}\right)^{-3} \times (81)^{-1} = 243, which simplifies to 39ร—3โˆ’4=2433^9 \times 3^{-4} = 243. This is a true statement, so we can verify that the solution m=โˆ’3m = -3 is correct.

Conclusion

In this article, we answered some common questions related to solving equations involving exponents. We discussed common mistakes to avoid, how to simplify terms, how to use exponent properties, real-world applications, and how to check your work. By understanding and applying exponent properties, we can solve complex equations and make informed decisions in various fields.