Problem 79 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Convert the Logarithmic Equations to Exponential Form

The logarithmic equations can be converted into equivalent exponential forms using the formula \( b = a^{\log_a(b)} \). This gives: 1. For \( \log_y {x} = 3 \), the equivalent exponential form is \( x = y^3 \).2. For \( \log_y {4x} = 5 \), the equivalent exponential form is \( 4x = y^5 \).

Step 2: Substitute and Solve for y

Now, from the first equation, \( x = y^3 \), the value of x can be substituted into the second equation, \( 4x = y^5 \). This gives: \( 4y^3 = y^5 \). Solving for y, we find that \( y = 0 \) or \( y = 2 \). The value \( y = 0 \) is not valid in the context of logarithmic equations as y is the base of the logarithm and it cannot be zero. So, \( y = 2 \) is the solution of the system for y.

Step 3: Substitute y back and Solve for x

Substitute \( y = 2 \) back into the first equation \( x = y^3 \), to get \( x = 2^3 = 8 \). So, \( x = 8 \) is the solution of the system for x.

Key Concepts

Exponential Form

In mathematics, converting logarithmic equations to exponential form is a handy tool. This process uses the fact that a logarithm is the inverse operation to exponentiation. Simply put, if you have a logarithmic equation like \( \log_b(a) = c \), it can be rewritten in its exponential form as \( a = b^c \). This conversion is like translating from one mathematical language to another.

By switching to exponential form:

• You gain a clearer view of the relationship between the numbers involved.
• This transformation often simplifies complex problems, making them easier to solve.
• It prepares the stage for further mathematical operations, such as solving systems of equations.

In our example, transforming \( \log_y{x} = 3 \) into \( x = y^3 \), provides a simple equation that we can solve directly once we know the value of \( y \). Similarly, \( \log_y{4x} = 5 \) becomes \( 4x = y^5 \). Such transformations are essential for unpacking deeper insights in mathematical problems.

Solve Systems of Equations

Systems of equations consist of multiple equations that need to be solved together. In the case of systems involving logarithms, transforming them into exponential form can greatly simplify your task.

Here’s why tackling systems of equations is important:

• You can find common solutions that satisfy all equations simultaneously, which reflects real-world interconnected problems.
• Understanding systems of equations enhances problem-solving skills, teaching you to manage multiple variables and constraints.
• Many complex problems in mathematics and other disciplines require solving systems of equations to uncover all hidden relationships.

With our problem, once we shift both logarithmic equations into their exponential forms \( x = y^3 \) and \( 4x = y^5 \), we can solve them together as a system. This lets us substitute values to find possible \( x \) and \( y \) that solve both equations at once.

Substitution Method

The substitution method is a technique used to solve systems of equations, by replacing one variable. It involves expressing one variable in terms of another using one of the equations, and substituting this expression into the other equations.

Why use the substitution method?

• It reduces the number of variables, turning a system into a single equation that is easier to manage.
• It’s particularly effective when one of the equations is already solved for one variable.
• Provides a step-by-step path, making it easier to track your work and ensure accuracy.

In our exercise, with \( x = y^3 \) from the first equation, we substitute \( x \) in the second equation to get \( 4y^3 = y^5 \). Simplifying this single equation allows us to solve for \( y \), yielding results we can use to find \( x \). The substitution method helps keep our calculations organized and focused on one unknown at a time.