Solve The Following System Of Logarithmic Equations:${ \begin{array}{l} \log_2 X^2 + \log_2 Y^3 = 1 \ \log X - \log_2 Y^2 = 4 \end{array} }$

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Introduction

Logarithmic equations are a fundamental concept in mathematics, and solving them requires a deep understanding of logarithmic properties and functions. In this article, we will focus on solving a system of logarithmic equations, which involves finding the values of variables that satisfy both equations simultaneously. We will use the given system of equations as an example and provide a step-by-step solution.

The System of Logarithmic Equations

The given system of logarithmic equations is:

{ \begin{array}{l} \log_2 x^2 + \log_2 y^3 = 1 \\ \log x - \log_2 y^2 = 4 \end{array} \}

Understanding Logarithmic Properties

Before we dive into solving the system of equations, it's essential to understand the properties of logarithms. The two main properties we will use are:

  • Product Property: log⁡b(xy)=log⁡bx+log⁡by\log_b (xy) = \log_b x + \log_b y
  • Quotient Property: log⁡b(xy)=log⁡bx−log⁡by\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y

Solving the First Equation

Let's start by solving the first equation:

log⁡2x2+log⁡2y3=1\log_2 x^2 + \log_2 y^3 = 1

Using the product property, we can rewrite the equation as:

log⁡2(x2y3)=1\log_2 (x^2y^3) = 1

Applying the Definition of Logarithm

The definition of logarithm states that if y=log⁡bxy = \log_b x, then by=xb^y = x. In this case, we can rewrite the equation as:

21=x2y32^1 = x^2y^3

Simplifying the equation, we get:

2=x2y32 = x^2y^3

Solving the Second Equation

Now, let's solve the second equation:

log⁡x−log⁡2y2=4\log x - \log_2 y^2 = 4

Using the quotient property, we can rewrite the equation as:

log⁡2(xy2)=4\log_2 \left(\frac{x}{y^2}\right) = 4

Applying the Definition of Logarithm

Using the definition of logarithm, we can rewrite the equation as:

24=xy22^4 = \frac{x}{y^2}

Simplifying the equation, we get:

16=xy216 = \frac{x}{y^2}

Solving for x and y

We now have two equations:

2=x2y32 = x^2y^3 16=xy216 = \frac{x}{y^2}

We can solve for x and y by substituting the second equation into the first equation:

2=(16y21)2y32 = \left(\frac{16y^2}{1}\right)^2y^3

Simplifying the equation, we get:

2=256y72 = 256y^7

Solving for y

Dividing both sides of the equation by 256, we get:

1128=y7\frac{1}{128} = y^7

Taking the seventh root of both sides, we get:

y=11287y = \sqrt[7]{\frac{1}{128}}

Simplifying the expression, we get:

y=12y = \frac{1}{2}

Solving for x

Now that we have the value of y, we can substitute it into one of the original equations to solve for x. Let's use the second equation:

16=xy216 = \frac{x}{y^2}

Substituting y = 1/2, we get:

16=x(12)216 = \frac{x}{\left(\frac{1}{2}\right)^2}

Simplifying the equation, we get:

16=x1416 = \frac{x}{\frac{1}{4}}

Multiplying both sides by 1/4, we get:

164=x\frac{16}{4} = x

Simplifying the expression, we get:

x=4x = 4

Conclusion

In this article, we solved a system of logarithmic equations using the product and quotient properties of logarithms. We first solved the first equation to get an expression for x2y3, and then used the second equation to get an expression for x/y^2. We then substituted the second equation into the first equation to get an equation in terms of y, and solved for y. Finally, we substituted the value of y into one of the original equations to solve for x. The final solution is x = 4 and y = 1/2.

Final Answer

The final answer is x = 4 and y = 1/2.

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Introduction

In our previous article, we solved a system of logarithmic equations using the product and quotient properties of logarithms. In this article, we will provide a Q&A guide to help you better understand the concepts and techniques involved in solving logarithmic equations.

Q: What are logarithmic equations?

A: Logarithmic equations are equations that involve logarithmic functions. They are used to solve problems that involve exponential growth or decay.

Q: What are the properties of logarithms?

A: The two main properties of logarithms are:

  • Product Property: log⁡b(xy)=log⁡bx+log⁡by\log_b (xy) = \log_b x + \log_b y
  • Quotient Property: log⁡b(xy)=log⁡bx−log⁡by\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y

Q: How do I solve a logarithmic equation?

A: To solve a logarithmic equation, you need to use the properties of logarithms to simplify the equation and isolate the variable. Here are the steps:

  1. Simplify the equation: Use the product and quotient properties to simplify the equation.
  2. Isolate the variable: Use algebraic manipulations to isolate the variable.
  3. Check the solution: Check the solution to make sure it satisfies the original equation.

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

A: A logarithmic equation is an equation that involves a logarithmic function, while an exponential equation is an equation that involves an exponential function. For example:

  • Logarithmic equation: log⁡2x=3\log_2 x = 3
  • Exponential equation: 2x=82^x = 8

Q: How do I solve a system of logarithmic equations?

A: To solve a system of logarithmic equations, you need to use the properties of logarithms to simplify the equations and isolate the variables. Here are the steps:

  1. Simplify the equations: Use the product and quotient properties to simplify the equations.
  2. Isolate the variables: Use algebraic manipulations to isolate the variables.
  3. Check the solution: Check the solution to make sure it satisfies both equations.

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

A: Here are some common mistakes to avoid when solving logarithmic equations:

  • Not using the correct properties of logarithms: Make sure to use the product and quotient properties correctly.
  • Not isolating the variable: Make sure to isolate the variable using algebraic manipulations.
  • Not checking the solution: Make sure to check the solution to make sure it satisfies the original equation.

Q: How do I check the solution to a logarithmic equation?

A: To check the solution to a logarithmic equation, you need to plug the solution back into the original equation and make sure it is true. Here are the steps:

  1. Plug in the solution: Plug the solution back into the original equation.
  2. Check if the equation is true: Check if the equation is true.
  3. Verify the solution: Verify that the solution satisfies the original equation.

Conclusion

In this article, we provided a Q&A guide to help you better understand the concepts and techniques involved in solving logarithmic equations. We covered topics such as the properties of logarithms, solving logarithmic equations, and checking the solution. We hope this guide has been helpful in your understanding of logarithmic equations.

Final Answer

The final answer is that logarithmic equations are a fundamental concept in mathematics, and solving them requires a deep understanding of logarithmic properties and functions. By following the steps outlined in this article, you can solve logarithmic equations and check your solutions to make sure they are true.