Which Expressions Are Equivalent To The One Below? Check All That Apply.$\ln \left(e^2\right$\]A. $2 \cdot \ln E$ B. 2 C. $2 E$ D. 1

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Introduction

In mathematics, equivalent expressions are those that have the same value or represent the same mathematical concept. When dealing with logarithmic and exponential functions, it's essential to understand the properties and rules that govern these functions. In this article, we will explore the equivalent expressions of the given expression ln⁑(e2)\ln \left(e^2\right) and identify the correct options.

Properties of Logarithmic and Exponential Functions

Before diving into the equivalent expressions, let's review the properties of logarithmic and exponential functions.

  • Exponential Function: The exponential function is defined as exe^x, where ee is a mathematical constant approximately equal to 2.71828. The exponential function has the following properties:
    • e0=1e^0 = 1
    • e1=ee^1 = e
    • exβ‹…ey=ex+ye^x \cdot e^y = e^{x+y}
  • Logarithmic Function: The logarithmic function is defined as ln⁑x\ln x, where xx is a positive real number. The logarithmic function has the following properties:
    • ln⁑1=0\ln 1 = 0
    • ln⁑e=1\ln e = 1
    • ln⁑xy=yβ‹…ln⁑x\ln x^y = y \cdot \ln x

Equivalent Expressions

Now, let's find the equivalent expressions of the given expression ln⁑(e2)\ln \left(e^2\right).

  • Option A: 2β‹…ln⁑e2 \cdot \ln e
    • Using the property ln⁑xy=yβ‹…ln⁑x\ln x^y = y \cdot \ln x, we can rewrite ln⁑(e2)\ln \left(e^2\right) as 2β‹…ln⁑e2 \cdot \ln e.
    • Since ln⁑e=1\ln e = 1, we can simplify 2β‹…ln⁑e2 \cdot \ln e to 22.
    • Therefore, option A is a correct equivalent expression.
  • Option B: 2
    • As we simplified option A, we can see that option B is also a correct equivalent expression.
  • Option C: 2e2 e
    • Using the property exβ‹…ey=ex+ye^x \cdot e^y = e^{x+y}, we can rewrite e2e^2 as eβ‹…ee \cdot e.
    • However, this does not help us find an equivalent expression for ln⁑(e2)\ln \left(e^2\right).
    • Therefore, option C is not a correct equivalent expression.
  • Option D: 1
    • Using the property ln⁑1=0\ln 1 = 0, we can see that ln⁑(e2)\ln \left(e^2\right) is not equal to 1.
    • Therefore, option D is not a correct equivalent expression.

Conclusion

In conclusion, the equivalent expressions of the given expression ln⁑(e2)\ln \left(e^2\right) are 2β‹…ln⁑e2 \cdot \ln e and 2. These expressions represent the same mathematical concept and have the same value.

Final Answer

The correct options are:

  • A. 2β‹…ln⁑e2 \cdot \ln e
  • B. 2

Additional Tips and Resources

  • To better understand the properties of logarithmic and exponential functions, we recommend reviewing the following resources:
    • Khan Academy: Logarithms and Exponents
    • Mathway: Logarithmic and Exponential Functions
    • Wolfram Alpha: Logarithmic and Exponential Functions

Introduction

In our previous article, we explored the equivalent expressions of the given expression ln⁑(e2)\ln \left(e^2\right) and identified the correct options. In this article, we will answer some frequently asked questions about logarithmic and exponential functions.

Q&A

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

A: A logarithmic function is the inverse of an exponential function. The logarithmic function ln⁑x\ln x is the inverse of the exponential function exe^x. In other words, ln⁑x\ln x asks the question "what power must we raise ee to, to get xx?", while exe^x asks the question "what number must we raise ee to, to get xx?".

Q: What is the value of ln⁑1\ln 1?

A: The value of ln⁑1\ln 1 is 0. This is because the logarithmic function ln⁑x\ln x is the inverse of the exponential function exe^x, and e0=1e^0 = 1.

Q: What is the value of ln⁑e\ln e?

A: The value of ln⁑e\ln e is 1. This is because the logarithmic function ln⁑x\ln x is the inverse of the exponential function exe^x, and e1=ee^1 = e.

Q: What is the property of logarithmic functions that states ln⁑xy=yβ‹…ln⁑x\ln x^y = y \cdot \ln x?

A: This property is known as the power rule of logarithms. It states that the logarithm of a power is equal to the exponent multiplied by the logarithm of the base.

Q: What is the property of exponential functions that states exβ‹…ey=ex+ye^x \cdot e^y = e^{x+y}?

A: This property is known as the product rule of exponents. It states that the product of two exponential functions with the same base is equal to the exponential function with the sum of the exponents.

Q: How do I evaluate the expression ln⁑(ex)\ln \left(e^x\right)?

A: To evaluate the expression ln⁑(ex)\ln \left(e^x\right), we can use the property ln⁑xy=yβ‹…ln⁑x\ln x^y = y \cdot \ln x. In this case, x=ex = e and y=xy = x, so we can rewrite the expression as xβ‹…ln⁑ex \cdot \ln e. Since ln⁑e=1\ln e = 1, we can simplify the expression to xx.

Q: How do I evaluate the expression ln⁑(ex)+ln⁑(ey)\ln \left(e^x\right) + \ln \left(e^y\right)?

A: To evaluate the expression ln⁑(ex)+ln⁑(ey)\ln \left(e^x\right) + \ln \left(e^y\right), we can use the property ln⁑xy=yβ‹…ln⁑x\ln x^y = y \cdot \ln x. In this case, we can rewrite the expression as ln⁑(ex+y)\ln \left(e^{x+y}\right). Using the property ln⁑xy=yβ‹…ln⁑x\ln x^y = y \cdot \ln x, we can simplify the expression to x+yx+y.

Conclusion

In conclusion, logarithmic and exponential functions are fundamental concepts in mathematics that have many applications in science, engineering, and finance. By understanding the properties and rules of these functions, we can evaluate expressions and solve problems with ease.

Final Tips and Resources

  • To better understand logarithmic and exponential functions, we recommend reviewing the following resources:
    • Khan Academy: Logarithms and Exponents
    • Mathway: Logarithmic and Exponential Functions
    • Wolfram Alpha: Logarithmic and Exponential Functions
  • Practice evaluating expressions and solving problems involving logarithmic and exponential functions to become proficient in these concepts.

By following these tips and resources, you can improve your understanding of logarithmic and exponential functions and become proficient in evaluating expressions and solving problems.