If Log ⁡ 10 Y + 3 Log ⁡ 10 X = 2 \log_{10} Y + 3 \log_{10} X = 2 Lo G 10 ​ Y + 3 Lo G 10 ​ X = 2 , Express Y Y Y In Terms Of X X X .

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Introduction

In this article, we will delve into the world of logarithms and explore a mathematical problem that requires us to express a variable in terms of another variable. The problem at hand is to express yy in terms of xx given the equation log10y+3log10x=2\log_{10} y + 3 \log_{10} x = 2. We will use various mathematical techniques and properties of logarithms to derive the expression for yy in terms of xx.

Understanding Logarithms

Before we proceed with the problem, let's take a moment to understand the concept of logarithms. A logarithm is the inverse operation of exponentiation. In other words, if ab=ca^b = c, then logac=b\log_a c = b. The logarithm of a number to a certain base is the exponent to which the base must be raised to produce that number. For example, log10100=2\log_{10} 100 = 2 because 102=10010^2 = 100.

Using Properties of Logarithms

To solve the problem, we will use the properties of logarithms. One of the most important properties of logarithms is the product rule, which states that loga(mn)=logam+logan\log_a (m \cdot n) = \log_a m + \log_a n. We will also use the power rule, which states that loga(mb)=blogam\log_a (m^b) = b \cdot \log_a m.

Deriving the Expression for y

Now that we have a good understanding of logarithms and their properties, let's proceed with the problem. We are given the equation log10y+3log10x=2\log_{10} y + 3 \log_{10} x = 2. Our goal is to express yy in terms of xx.

Using the power rule, we can rewrite the equation as log10y+log10x3=2\log_{10} y + \log_{10} x^3 = 2. This simplifies to log10y+log10(x3)=2\log_{10} y + \log_{10} (x^3) = 2.

Next, we can use the product rule to combine the two logarithms on the left-hand side of the equation. This gives us log10(yx3)=2\log_{10} (y \cdot x^3) = 2.

Now, we can use the definition of logarithms to rewrite the equation in exponential form. This gives us yx3=102y \cdot x^3 = 10^2.

Simplifying the right-hand side of the equation, we get yx3=100y \cdot x^3 = 100.

Finally, we can solve for yy by dividing both sides of the equation by x3x^3. This gives us y=100x3y = \frac{100}{x^3}.

Conclusion

In this article, we have derived the expression for yy in terms of xx given the equation log10y+3log10x=2\log_{10} y + 3 \log_{10} x = 2. We used various mathematical techniques and properties of logarithms to solve the problem. The final expression for yy in terms of xx is y=100x3y = \frac{100}{x^3}.

Example Use Cases

The expression for yy in terms of xx can be used in a variety of mathematical and real-world applications. For example, it can be used to model the relationship between two variables in a scientific experiment or to solve a system of equations.

Step-by-Step Solution

Here is a step-by-step solution to the problem:

  1. Use the power rule to rewrite the equation as log10y+log10x3=2\log_{10} y + \log_{10} x^3 = 2.
  2. Use the product rule to combine the two logarithms on the left-hand side of the equation. This gives us log10(yx3)=2\log_{10} (y \cdot x^3) = 2.
  3. Use the definition of logarithms to rewrite the equation in exponential form. This gives us yx3=102y \cdot x^3 = 10^2.
  4. Simplify the right-hand side of the equation to get yx3=100y \cdot x^3 = 100.
  5. Solve for yy by dividing both sides of the equation by x3x^3. This gives us y=100x3y = \frac{100}{x^3}.

Mathematical Derivations

Here are the mathematical derivations for the properties of logarithms used in the solution:

  • Product Rule: loga(mn)=logam+logan\log_a (m \cdot n) = \log_a m + \log_a n
  • Power Rule: loga(mb)=blogam\log_a (m^b) = b \cdot \log_a m

Real-World Applications

The expression for yy in terms of xx can be used in a variety of real-world applications, such as:

  • Scientific Experiments: The expression can be used to model the relationship between two variables in a scientific experiment.
  • Engineering: The expression can be used to solve a system of equations in engineering applications.
  • Economics: The expression can be used to model the relationship between two variables in economic models.

Conclusion

Introduction

In our previous article, we derived the expression for yy in terms of xx given the equation log10y+3log10x=2\log_{10} y + 3 \log_{10} x = 2. We used various mathematical techniques and properties of logarithms to solve the problem. In this article, we will answer some of the most frequently asked questions related to the problem.

Q: What is the final expression for y in terms of x?

A: The final expression for yy in terms of xx is y=100x3y = \frac{100}{x^3}.

Q: How did you derive the expression for y in terms of x?

A: We used the power rule and the product rule of logarithms to derive the expression for yy in terms of xx. We started with the equation log10y+3log10x=2\log_{10} y + 3 \log_{10} x = 2 and used the power rule to rewrite it as log10y+log10x3=2\log_{10} y + \log_{10} x^3 = 2. We then used the product rule to combine the two logarithms on the left-hand side of the equation, which gave us log10(yx3)=2\log_{10} (y \cdot x^3) = 2. We then used the definition of logarithms to rewrite the equation in exponential form, which gave us yx3=102y \cdot x^3 = 10^2. We then simplified the right-hand side of the equation to get yx3=100y \cdot x^3 = 100. Finally, we solved for yy by dividing both sides of the equation by x3x^3, which gave us y=100x3y = \frac{100}{x^3}.

Q: What are some of the real-world applications of the expression for y in terms of x?

A: The expression for yy in terms of xx can be used in a variety of real-world applications, such as:

  • Scientific Experiments: The expression can be used to model the relationship between two variables in a scientific experiment.
  • Engineering: The expression can be used to solve a system of equations in engineering applications.
  • Economics: The expression can be used to model the relationship between two variables in economic models.

Q: What are some of the mathematical techniques used to derive the expression for y in terms of x?

A: We used the following mathematical techniques to derive the expression for yy in terms of xx:

  • Power Rule: loga(mb)=blogam\log_a (m^b) = b \cdot \log_a m
  • Product Rule: loga(mn)=logam+logan\log_a (m \cdot n) = \log_a m + \log_a n
  • Definition of Logarithms: ab=c    logac=ba^b = c \implies \log_a c = b

Q: Can you provide a step-by-step solution to the problem?

A: Here is a step-by-step solution to the problem:

  1. Use the power rule to rewrite the equation as log10y+log10x3=2\log_{10} y + \log_{10} x^3 = 2.
  2. Use the product rule to combine the two logarithms on the left-hand side of the equation. This gives us log10(yx3)=2\log_{10} (y \cdot x^3) = 2.
  3. Use the definition of logarithms to rewrite the equation in exponential form. This gives us yx3=102y \cdot x^3 = 10^2.
  4. Simplify the right-hand side of the equation to get yx3=100y \cdot x^3 = 100.
  5. Solve for yy by dividing both sides of the equation by x3x^3. This gives us y=100x3y = \frac{100}{x^3}.

Q: What are some of the common mistakes to avoid when deriving the expression for y in terms of x?

A: Some of the common mistakes to avoid when deriving the expression for yy in terms of xx include:

  • Forgetting to use the power rule: Make sure to use the power rule to rewrite the equation as log10y+log10x3=2\log_{10} y + \log_{10} x^3 = 2.
  • Forgetting to use the product rule: Make sure to use the product rule to combine the two logarithms on the left-hand side of the equation.
  • Forgetting to simplify the right-hand side of the equation: Make sure to simplify the right-hand side of the equation to get yx3=100y \cdot x^3 = 100.
  • Forgetting to solve for y: Make sure to solve for yy by dividing both sides of the equation by x3x^3.

Conclusion

In this article, we have answered some of the most frequently asked questions related to the problem of expressing yy in terms of xx given the equation log10y+3log10x=2\log_{10} y + 3 \log_{10} x = 2. We have provided a step-by-step solution to the problem and highlighted some of the common mistakes to avoid when deriving the expression for yy in terms of xx.