Solve: ( 1 5 , 000 ) − 2 Z ⋅ 5 , 000 − 2 Z + 2 = 5 , 000 \left(\frac{1}{5,000}\right)^{-2z} \cdot 5,000^{-2z+2} = 5,000 ( 5 , 000 1 ​ ) − 2 Z ⋅ 5 , 00 0 − 2 Z + 2 = 5 , 000 A. Z = − 1 Z = -1 Z = − 1 B. Z = 0 Z = 0 Z = 0 C. Z = 1 Z = 1 Z = 1 D. No Solution

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Introduction

Exponential equations are a fundamental concept in mathematics, and solving them requires a deep understanding of algebraic manipulations and properties of exponents. In this article, we will focus on solving a specific type of exponential equation, which involves a product of two exponential expressions. We will use the given equation (15,000)−2z⋅5,000−2z+2=5,000\left(\frac{1}{5,000}\right)^{-2z} \cdot 5,000^{-2z+2} = 5,000 as a case study to demonstrate the steps involved in solving such equations.

Understanding the Equation

The given equation is (15,000)−2z⋅5,000−2z+2=5,000\left(\frac{1}{5,000}\right)^{-2z} \cdot 5,000^{-2z+2} = 5,000. To begin solving this equation, we need to understand the properties of exponents and how they interact with each other. The equation involves two exponential expressions, (15,000)−2z\left(\frac{1}{5,000}\right)^{-2z} and 5,000−2z+25,000^{-2z+2}, which are multiplied together to equal 5,0005,000.

Simplifying the Equation

To simplify the equation, we can start by using the property of exponents that states am⋅an=am+na^m \cdot a^n = a^{m+n}. We can apply this property to the given equation by combining the two exponential expressions:

(15,000)−2z⋅5,000−2z+2=(15,000)−2z⋅(15,000)−2z+2⋅5,0002\left(\frac{1}{5,000}\right)^{-2z} \cdot 5,000^{-2z+2} = \left(\frac{1}{5,000}\right)^{-2z} \cdot \left(\frac{1}{5,000}\right)^{-2z+2} \cdot 5,000^2

Using the property of exponents that states am⋅an=am+na^m \cdot a^n = a^{m+n}, we can simplify the equation further:

(15,000)−2z⋅(15,000)−2z+2⋅5,0002=(15,000)−4z+2⋅5,0002\left(\frac{1}{5,000}\right)^{-2z} \cdot \left(\frac{1}{5,000}\right)^{-2z+2} \cdot 5,000^2 = \left(\frac{1}{5,000}\right)^{-4z+2} \cdot 5,000^2

Using the Property of Exponents

Now that we have simplified the equation, we can use the property of exponents that states am⋅an=am+na^m \cdot a^n = a^{m+n} to rewrite the equation:

(15,000)−4z+2⋅5,0002=(15,000)−4z+2⋅(15,000)−2⋅5,0002\left(\frac{1}{5,000}\right)^{-4z+2} \cdot 5,000^2 = \left(\frac{1}{5,000}\right)^{-4z+2} \cdot \left(\frac{1}{5,000}\right)^{-2} \cdot 5,000^2

Using the property of exponents that states am⋅an=am+na^m \cdot a^n = a^{m+n}, we can simplify the equation further:

(15,000)−4z+2⋅(15,000)−2⋅5,0002=(15,000)−4z+4⋅5,0002\left(\frac{1}{5,000}\right)^{-4z+2} \cdot \left(\frac{1}{5,000}\right)^{-2} \cdot 5,000^2 = \left(\frac{1}{5,000}\right)^{-4z+4} \cdot 5,000^2

Solving for z

Now that we have simplified the equation, we can solve for zz. We can start by using the property of exponents that states am=ana^m = a^n implies m=nm = n. We can apply this property to the given equation by setting the exponents equal to each other:

−4z+4=0-4z+4 = 0

Solving for zz, we get:

z=1z = 1

Conclusion

In this article, we have demonstrated the steps involved in solving a specific type of exponential equation. We have used the given equation (15,000)−2z⋅5,000−2z+2=5,000\left(\frac{1}{5,000}\right)^{-2z} \cdot 5,000^{-2z+2} = 5,000 as a case study to illustrate the process of simplifying and solving the equation. We have shown that the solution to the equation is z=1z = 1. This demonstrates the importance of understanding the properties of exponents and how they interact with each other in solving exponential equations.

Final Answer

The final answer to the equation (15,000)−2z⋅5,000−2z+2=5,000\left(\frac{1}{5,000}\right)^{-2z} \cdot 5,000^{-2z+2} = 5,000 is:

  • z=1z = 1

Introduction

In our previous article, we demonstrated the steps involved in solving a specific type of exponential equation. We used the equation (15,000)−2z⋅5,000−2z+2=5,000\left(\frac{1}{5,000}\right)^{-2z} \cdot 5,000^{-2z+2} = 5,000 as a case study to illustrate the process of simplifying and solving the equation. In this article, we will provide a Q&A guide to help you better understand the concepts and techniques involved in solving exponential equations.

Q: What is an exponential equation?

A: An exponential equation is an equation that involves a variable in the exponent of a number. For example, the equation 2x=82^x = 8 is an exponential equation because the variable xx is in the exponent of the number 22.

Q: What are the properties of exponents?

A: The properties of exponents are a set of rules that govern how exponents interact with each other. Some of the key properties of exponents include:

  • am⋅an=am+na^m \cdot a^n = a^{m+n}
  • am⋅bn=am⋅bna^m \cdot b^n = a^m \cdot b^n
  • (am)n=amn(a^m)^n = a^{mn}
  • a−m=1ama^{-m} = \frac{1}{a^m}

Q: How do I simplify an exponential equation?

A: To simplify an exponential equation, you can use the properties of exponents to combine the exponents and simplify the equation. For example, if you have the equation 2x⋅2y=2x+y2^x \cdot 2^y = 2^{x+y}, you can simplify it by combining the exponents.

Q: How do I solve an exponential equation?

A: To solve an exponential equation, you can use the properties of exponents to isolate the variable and solve for its value. For example, if you have the equation 2x=82^x = 8, you can solve for xx by using the property of exponents that states am=ana^m = a^n implies m=nm = n.

Q: What is the difference between an exponential equation and a logarithmic equation?

A: An exponential equation is an equation that involves a variable in the exponent of a number, while a logarithmic equation is an equation that involves a variable as the exponent of a number. For example, the equation 2x=82^x = 8 is an exponential equation, while the equation x=log⁡28x = \log_2 8 is a logarithmic equation.

Q: How do I use logarithms to solve an exponential equation?

A: To use logarithms to solve an exponential equation, you can take the logarithm of both sides of the equation and use the properties of logarithms to simplify and solve for the variable. For example, if you have the equation 2x=82^x = 8, you can take the logarithm of both sides and use the property of logarithms that states log⁡ab=blog⁡a\log a^b = b \log a to simplify and solve for xx.

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

A: Some common mistakes to avoid when solving exponential equations include:

  • Not using the properties of exponents correctly
  • Not isolating the variable correctly
  • Not checking the domain of the equation
  • Not using logarithms to solve the equation when necessary

Conclusion

In this article, we have provided a Q&A guide to help you better understand the concepts and techniques involved in solving exponential equations. We have covered topics such as the properties of exponents, simplifying exponential equations, solving exponential equations, and using logarithms to solve exponential equations. By following the tips and techniques outlined in this article, you should be able to solve exponential equations with confidence.

Final Tips

  • Always use the properties of exponents correctly
  • Always isolate the variable correctly
  • Always check the domain of the equation
  • Always use logarithms to solve the equation when necessary

By following these tips and techniques, you should be able to solve exponential equations with ease and confidence.