Solving The System Of Equations Xy = 20 And X^(lg√x) = 2
In this article, we will delve into the fascinating world of solving systems of equations. Specifically, we will tackle the system:
\begin{cases}
xy = 20, \\
x^{\lg \sqrt{x}} = 2
\end{cases}
This system combines algebraic and logarithmic equations, presenting a unique challenge that requires a blend of techniques to solve. We will systematically break down the problem, applying relevant logarithmic and algebraic properties to arrive at the solution. This exploration will not only provide a step-by-step solution to this particular system but also enhance your understanding of how to approach similar mathematical problems.
Before we jump into the solution, let's take a closer look at the two equations we are dealing with. The first equation, xy = 20, is a straightforward algebraic equation that establishes a relationship between two variables, x and y. It tells us that the product of x and y is equal to 20. This equation is relatively simple and provides a direct link between the two variables, which we will exploit later in our solution.
The second equation, x^(lg √x) = 2, is where things get more interesting. This equation involves a variable, x, raised to a power that includes a logarithm. Specifically, the exponent is lg √x, where lg denotes the base-10 logarithm (logarithm with base 10). The square root of x inside the logarithm adds another layer of complexity. To tackle this equation, we will need to utilize the properties of logarithms and exponents, such as the power rule of logarithms and the relationship between logarithms and exponents. The presence of the logarithm suggests that we might need to consider the domain of the logarithmic function, ensuring that our solutions for x are valid (i.e., x must be positive for the logarithm to be defined). This equation is the key to unlocking the system, as it will allow us to isolate x and subsequently find the corresponding value of y using the first equation.
Now, let's embark on the journey of solving the system. Our approach will involve manipulating the equations, applying logarithmic properties, and using substitution to arrive at the solutions for x and y. Each step will be carefully explained, providing a clear understanding of the reasoning behind it.
Step 1: Manipulating the Second Equation
Our primary focus will be on the second equation, x^(lg √x) = 2, as it contains the logarithmic term and is more complex. To simplify this equation, we'll employ the power rule of logarithms. However, before we do that, let's rewrite the square root as an exponent: √x = x^(1/2). Substituting this into the exponent, we get:
lg √x = lg x^(1/2)
Now, we can apply the power rule of logarithms, which states that lg a^b = b lg a. Applying this rule, we have:
lg x^(1/2) = (1/2) lg x
Substituting this back into the original equation's exponent, we get:
x^((1/2) lg x) = 2
This form is slightly more manageable. To further simplify, we'll take the base-10 logarithm (lg) of both sides of the equation. This is a crucial step, as it allows us to bring the exponent down using the power rule of logarithms. Applying the logarithm to both sides, we get:
lg(x^((1/2) lg x)) = lg 2
Now, we apply the power rule of logarithms again to the left side:
((1/2) lg x) lg x = lg 2
This simplifies to:
(1/2) (lg x)^2 = lg 2
Step 2: Isolating (lg x)^2
To isolate the term (lg x)^2, we multiply both sides of the equation by 2:
(lg x)^2 = 2 lg 2
This equation now expresses the square of the logarithm of x in terms of a constant. This is a significant step forward, as it allows us to solve for lg x.
Step 3: Solving for lg x
To solve for lg x, we take the square root of both sides of the equation. Remember that taking the square root introduces both positive and negative solutions:
lg x = ±√(2 lg 2)
This gives us two possible values for lg x: a positive value and a negative value. We will consider each case separately.
Step 4: Finding x
To find the values of x, we need to undo the logarithm. Since we are using the base-10 logarithm, we can do this by raising 10 to the power of both sides of the equation. For the positive case:
lg x = √(2 lg 2)
x = 10^(√(2 lg 2))
And for the negative case:
lg x = -√(2 lg 2)
x = 10^(-√(2 lg 2))
These are the two possible values for x. Let's denote them as x₁ and x₂:
x₁ = 10^(√(2 lg 2))
x₂ = 10^(-√(2 lg 2))
Step 5: Finding y
Now that we have the values for x, we can use the first equation, xy = 20, to find the corresponding values for y. For each value of x, we can solve for y by dividing 20 by x.
For x₁:
y₁ = 20 / x₁ = 20 / (10^(√(2 lg 2)))
For x₂:
y₂ = 20 / x₂ = 20 / (10^(-√(2 lg 2)))
Thus, we have found the two pairs of solutions for the system of equations.
The solutions to the system of equations are:
(x₁, y₁) = (10^(√(2 lg 2)), 20 / (10^(√(2 lg 2))))
(x₂, y₂) = (10^(-√(2 lg 2)), 20 / (10^(-√(2 lg 2))))
These are the two pairs of values for x and y that satisfy both equations in the system.
To ensure that our solutions are correct, we can substitute them back into the original equations and verify that they hold true. This is a crucial step in problem-solving, as it helps catch any potential errors made during the process.
Verifying Solution 1: (x₁, y₁)
Let's substitute x₁ = 10^(√(2 lg 2)) and y₁ = 20 / (10^(√(2 lg 2))) into the original equations.
Equation 1: xy = 20
x₁ y₁ = (10^(√(2 lg 2))) * (20 / (10^(√(2 lg 2)))) = 20
The first equation is satisfied.
Equation 2: x^(lg √x) = 2
Substituting x₁ into the second equation, we have:
(10^(√(2 lg 2)))^(lg √(10^(√(2 lg 2))))
Let's simplify the exponent:
lg √(10^(√(2 lg 2))) = lg (10^(√(2 lg 2)/2)) = (√(2 lg 2)/2) * lg 10 = √(lg 2 / 2)
Now, the equation becomes:
(10^(√(2 lg 2)))^(√(lg 2 / 2))
Using the property (ab)c = a^(b*c), we get:
10^(√(2 lg 2) * √(lg 2 / 2)) = 10^(√(2 lg 2 * lg 2 / 2)) = 10^(lg 2) = 2
The second equation is also satisfied. Therefore, the solution (x₁, y₁) is verified.
Verifying Solution 2: (x₂, y₂)
Now, let's substitute x₂ = 10^(-√(2 lg 2)) and y₂ = 20 / (10^(-√(2 lg 2))) into the original equations.
Equation 1: xy = 20
x₂ y₂ = (10^(-√(2 lg 2))) * (20 / (10^(-√(2 lg 2)))) = 20
The first equation is satisfied.
Equation 2: x^(lg √x) = 2
Substituting x₂ into the second equation, we have:
(10^(-√(2 lg 2)))^(lg √(10^(-√(2 lg 2))))
Let's simplify the exponent:
lg √(10^(-√(2 lg 2))) = lg (10^(-√(2 lg 2)/2)) = (-√(2 lg 2)/2) * lg 10 = -√(lg 2 / 2)
Now, the equation becomes:
(10^(-√(2 lg 2)))^(-√(lg 2 / 2))
Using the property (ab)c = a^(b*c), we get:
10^(-√(2 lg 2) * -√(lg 2 / 2)) = 10^(√(2 lg 2 * lg 2 / 2)) = 10^(lg 2) = 2
The second equation is also satisfied. Therefore, the solution (x₂, y₂) is verified.
In this article, we have successfully solved the system of equations:
\begin{cases}
xy = 20, \\
x^{\lg \sqrt{x}} = 2
\end{cases}
We achieved this by systematically applying logarithmic and algebraic properties, including the power rule of logarithms and the relationship between logarithms and exponents. The solutions we found are:
(x₁, y₁) = (10^(√(2 lg 2)), 20 / (10^(√(2 lg 2))))
(x₂, y₂) = (10^(-√(2 lg 2)), 20 / (10^(-√(2 lg 2))))
We also verified these solutions by substituting them back into the original equations, ensuring their correctness. This problem highlights the importance of understanding and applying mathematical properties to solve complex equations. The process of solving this system not only provides the answers but also enhances our problem-solving skills and deepens our understanding of the interplay between algebraic and logarithmic functions. The techniques used here can be applied to a wide range of similar mathematical problems, making this a valuable exercise in mathematical reasoning and problem-solving.
