Problem 80 - College Algebra Essentials

Question

Solve the systems in Exercises \(79-80 .\) $$ \left\\{\begin{array}{l} \log x^{2}=y+3 \\ \log x=y-1 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 10000, y = 5\).

Step 1: Express \(x\) in terms of \(y\)

From the second equation we could easily express x in terms of \(y\) by removing the logarithm, that is using the inverse of a logarithm, an exponentiation. The equation \(\log x=y-1\) becomes \(x=10^{y-1}\).

Step 2: Substituting \(x\)

Substitute \(x=10^{y-1}\) into the first equation of the original system. After doing that, it will look like this: \(\log (10^{y-1})^{2}=y+3\). Simplify this equation to: \(2y-2 = y+3\).

Step 3: Solving for \(y\)

Solve the equation in Step 2 for \(y\). You can do this by first subtracting \(y\) from both sides to get \(y - 2 = 3\), and then adding 2 to both sides to get \(y = 5.\)

Step 4: Solving for \(x\)

Solve for \(x\) using the equation from Step 1 to substitute the value of \(y\), which leads to \(x = 10^{5-1}\). Simplify this to get \(x = 10000\).

Key Concepts

Logarithmic Equations

Logarithmic equations form an integral part of algebra and understanding them enables us to handle complex systems of equations. A logarithmic equation is one where the variable is inside a logarithm. In this exercise, we have

• \( \log x^2 = y + 3 \)
• \( \log x = y - 1 \)

Understanding how to manipulate these expressions involves using properties of logarithms such as the power rule, \( \log a^b = b \cdot \log a \), which we leverage to simplify and solve.
Dealing with logarithmic equations might initially seem daunting, but recognizing them means knowing how to apply their inverse operation, exponentiation, to simplify and resolve the system.

Exponentiation

Exponentiation is the process of raising a number, known as the base, to the power of an exponent. It is the inverse operation of taking logarithms. In the given system of equations, to solve \( \log x = y - 1 \), we use exponentiation:

• The logarithm base here is 10. Thus, \( x = 10^{y-1} \).

This conversion from logarithmic to exponential form is crucial as it takes the variable out of the logarithm, setting up the equation for straightforward algebraic manipulation.
By applying exponentiation, you simplify logarithmic equations, making them easier to solve and understand. It transforms a complex logarithmic form into a simpler equation readily understood in algebraic terms.

Solving for Variables

After transforming logarithmic expressions, you often need to solve for the variable. For this specific problem, we wanted to find the values of both \( x \) and \( y \).

• From \( x = 10^{y-1} \), substituting into the other equation \( \log (10^{y-1})^2 = y + 3 \), simplifies to \( 2(y-1) = y + 3 \).
• To solve for \( y \), isolate \( y \) by rearranging the equation: solve \( 2y - 2 = y + 3 \), which further simplifies to \( y = 5 \).

Now that the value for \( y \) is found, substitute back to find \( x \) from the equation \( x = 10^{y-1} \).
This sequence of solving for variables is essential in systems of equations, simplifying one equation to express a variable in terms of another, substituting it back, and simplifying to find precise solutions.

Algebraic Manipulation

Algebraic manipulation involves rearranging an equation to isolate variables and solve systems of equations. In solving our system, it plays a vital role:

• By applying exponent rules to expressions like \( (10^{y-1})^2 \), simplifying becomes \( 10^{2y-2} \). The algebraic manipulation rules that bring these expressions to simpler forms are key here.
• Through subtraction, addition, and rearranging, finding \( y = 5 \) and subsequently \( x = 10000 \).

Effective algebraic manipulation starts with understanding properties of numbers and operations, allowing one to transform and solve even seemingly complex relationships with ease.
This skill unlocks the ability to tackle broader mathematical problems, bridging the understanding of simple operations to more complex systems of equations. It centers on making efficient and logical steps to simplify and resolve equations.