Problem 79 - College Algebra Essentials

Question

Solve the systems in Exercises \(79-80 .\) $$ \left\\{\begin{array}{l} \log _{y} x=3 \\ \log _{y}(4 x)=5 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 8\), \(y = 2\) and \(x = -8\), \(y = -2\).

Step 1: Transform the Log Equations to Exponential Equations

Take the log equations and convert them into their exponential forms. This gives us two equations: \[y^3 = x\] and \[y^5 = 4x\]

Step 2: Express both equations in terms of 'y'

Express the equation \(y^3 = x\) as \(x = y^3 \), then substitute \(x\) in the second equation \(y^5 = 4x\), resulting in \(y^5 = 4y^3\).

Step 3: Solve for 'y'

Solve the equation \(y^5 = 4y^3\). This equation simplifies to \(y^5 - 4y^3 = 0\). Factor out \(y^3\) to get \(y^3(y^2 - 4) = 0\). Hence, \(y = 0\), \(y^2 - 4 = 0\), or \(y = \sqrt{4}\). This gives us two values for 'y', which are \(y = 0, 2\) and \(y = -2\). However, 'log0' is undefined in mathematics. Thus, \(y = 0\) is an invalid solution.

Step 4: Solve for 'x'

Substitute \(y = 2\) and \(y = -2\) into the first equation \(x = y^3 = 2^3 = 8\) and \(x = (-2)^3 = -8\).

Key Concepts

Logarithmic Equations

A logarithmic equation is one in which the unknown variable appears within a logarithm, making the properties of logarithms essential for solving it. Logarithmic equations are often used to solve problems involving exponential growth or decay, such as in population studies or radioactive decay.

For instance, the problem given is a system of two logarithmic equations: \(\log_{y} x = 3\) and \(\log_{y}(4x) = 5\). To solve these, we need to understand how to manipulate logarithms. A fundamental property states that \(\log_b a = c\) implies \(b^c = a\). This property allows us to convert logarithmic equations into exponential form, which is more straightforward to solve algebraically.

• Translating to Exponential Form: The given equations translate to \(y^3 = x\) and \(y^5 = 4x\), respectively, using the property \(\log_b a = c \implies b^c = a\).
• Understanding the Base: In our exercise, 'y' is the base of the logarithm, which must be positive and not equal to 1.
• Dealing with Multiple Logarithmic Equations: When solving systems with more than one logarithmic equation, it's often necessary to express one equation in terms of the other, as demonstrated in the exercise.

Exponential Form

Exponential form is a way of representing numbers that makes use of exponents. An exponent indicates how many times the base, which is the number being multiplied, is used as a factor. It's represented as \(b^n\), where 'b' is the base, and 'n' is the exponent.

When we convert a logarithmic equation like \(\log_{y} x = 3\) into its exponential form, it's built upon the knowledge that the two are inverse operations. The exponential equivalent is \(x = y^3\). This conversion is the critical step that bridges the gap between the logarithmic form, which can seem abstract, and the more intuition-friendly exponential form.

• Application: We applied this conversion in our system of equations by transforming \(\log_{y} x = 3\) to \(x = y^3\) and \(\log_{y}(4x) = 5\) to \(y^5 = 4x\). By doing so, we're preparing the equations for algebraic manipulation.
• Importance: This method not only simplifies the problem but also highlights the relationship between exponents and logarithms, which is crucial for understanding more complex mathematical concepts.

Algebraic Factoring

Algebraic factoring is a powerful tool for simplifying polynomial equations and finding their roots. It involves breaking down a complex expression into simpler factors that, when multiplied together, give the original expression.

In the context of the exercise, we factored the equation \(y^5 - 4y^3 = 0\) after substituting \(x\) with \(y^3\). The technique used here is factoring by grouping, which allows us to take a common factor out. In this case, \(y^3\) is the common factor across both terms, and we can express the equation as \(y^3(y^2 - 4) = 0\).

• Zero Product Property: Once factored, we apply the zero product property, which states if a product of factors equals zero, then at least one of the factors must be zero. This leads to \(y = 0\) or \(y^2 - 4 = 0\).
• Further Factoring: The second factor \(y^2 - 4\) can be recognized as a difference of squares and factored further to \(y - 2)(y + 2) = 0\), yielding potential solutions for 'y'.

It's important to note that when solving logarithmic equations, we must also consider the domain restrictions of our solutions. Since logarithms are undefined for non-positive numbers, any value of 'y' that would result in a log of zero or a negative number must be excluded from the solution set.