Problem 80 - College Algebra

Question

Solve the systems $$\left\\{\begin{array}{l} \log x^{2}-y+3 \\ \log x-y-1 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 0.01\) and \(y = 0\).

Step 1: Rearrange the equations

The equations are given as: \(\log x^{2}-y+3=0\) and \(\log x-y-1=0\). First, we can rearrange both equations to isolate 'y' (since there is a single 'y' in both equations). We end up with: \(y = \log x^{2}+3\) and \(y = \log x + 1\).

Step 2: Equate the two expression for y

Since both expressions are equal to 'y', we can equate them: \(\log x^{2}+3 = \log x + 1\).

Step 3: Simplify the equation

Subtract \(\log x\) and 1 from both sides to obtain \(\log x^{2} - \log x + 2 = 0\). Using the properties of logarithms, specifically the quotient rule: \(\log a - \log b = \log (a/b)\), this equation becomes \(\log (\frac{x^{2}}{x}) + 2 = 0\), which simplifies to \(\log x + 2 = 0\). From this, \(x = 10^{-2}\), that is, \(x = 0.01\).

Step 4: Substitute x into one of the first rearranged equations

Substitute \(x = 0.01\) into the equation \(y = \log x + 1\) to solve for \(y\). Hence, \(y = \log 0.01 + 1 = -1 + 1 = 0\).

Key Concepts

Logarithmic Equations

Logarithmic equations are equations that involve logarithms of unknown variables. Understanding how to solve such equations is crucial for tackling problems related to exponential growth and decay.
In the given exercise, both equations contain logarithms with variable "x."
By recognizing this structure, we can approach the solution by simplifying and isolating the variable.

• These equations require traditional algebraic techniques, often combined with logarithm rules, like moving terms to different sides of the equation.
• These rules help convert complex logarithmic forms into simpler equations that are easier to handle.

In our solution, notice the use of inverse operations to eliminate terms, giving us cleaner expressions to work from. By applying logarithm properties, we can transform equations and solve for the unknown variables efficiently.

Substitution Method

The substitution method is an approach used to solve systems of equations, whether they involve polynomials or logarithms.
This method is especially useful when isolating one variable in the equations. After rearranging each equation from the system, we write both in terms of variable "y."
The logic is straightforward: if "y" equals two different expressions, those expressions must be equal to each other. In simpler terms, it's like saying two sides of an equation are equivalent because they represent the same value for "y."

• First, each equation is rearranged to make one variable ("y" in this instance) the subject.
• Then, equations are set equal to each other. Now, with a single equation involving only one variable ("x"), solving becomes much simpler.

Once you solve for one of the variables, back-substitution allows you to find the other variable, completing the solution of the system.

Logarithm Properties

Logarithm properties play an essential role in simplifying logarithmic equations. These include:

• Product Property: \(\log(a \cdot b) = \log a + \log b\)
• Quotient Property: \(\log(a / b) = \log a - \log b\)
• Power Property: \(\log(a^b) = b \cdot \log a\)

In our solution, the quotient property turns two logs subtracted into a single logarithm divided. This is crucial because it allows us to work with simpler equations.
For example, in the equation \(\log x^2 - \log x = \log(x^2/x)\), simplifying into \(\log x\) reduced complexity immediately.
Understanding these properties allows one to convert and solve any seemingly difficult logarithmic expression, as seen in the exercise.