Problem 64 - College Algebra

Question

Solve the systems $$\begin{aligned} &\log _{y} x=3\\\ &\log _{y}(4 x)=5 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions of the system are: \(x=8\) and \(y=2\).

Step 1: Change the form of Equations

To start, let's convert both logarithmic equations into exponential form to simplify them. This results in the following equations: Equation 1: \(y^3 = x\), Equation 2: \(y^5 = 4x\).

Step 2: Substitute Equation to solve for \(y\)

From Equation 1, we isolate \(x\) to be \(y^3\). Substituting this into Equation 2 gives us \(y^5 = 4y^3\). Subtracting \(4y^3\) from both sides forms a new equation: \(y^5 - 4y^3 = 0\). Factor out \(y^3\), contributing to \(y^3(y^2 - 4) = 0\). Therefore \(y\) can be \(0\) or equal to \(±2\). Substituting \(-2\) into Equation 1 gives an invalid result as the base of a log cannot be negative. So, \(y\) can either be \(0\) or \(2\).

Step 3: Substitute \(y\) into Equation 1 to find \(x\)

Substituting \(y=2\) into Equation 1 gives \(2^3 = x\), so \(x = 8\). When \(y=0\), this produces an invalid result in Equation 1 since the base of a log can't be zero. Therefore, the only solution for the system of equations is \(y = 2\) and \(x = 8\).

Key Concepts

Understanding the System of Equations

A system of equations is a set of two or more equations with common variables. In this exercise, we are given a system of logarithmic equations where both equations involve the variables \(x\) and \(y\). To solve such a system, the goal is to determine the values of the variables that satisfy all the equations simultaneously.

Here, we have:

• \(\log_{y} x = 3\)
• \(\log_{y}(4x) = 5\)

The key approach is to transform these into a form that is easier to solve. Using properties of logarithms often helps, but in this case, converting into exponential form is more direct.

Exponential Form Transformation

To make logarithmic equations easier to handle, we convert them into their exponential form. This form unveils the relationships between the variables more clearly.

Given a logarithmic equation \(\log_b a = c\), the equivalent exponential form is \(b^c = a\). Applying this to our system of equations:

• From \(\log_{y} x = 3\), we get \(x = y^3\).
• From \(\log_{y}(4x) = 5\), we obtain \(4x = y^5\).

Now, instead of dealing with logarithms, we work with powers of \(y\), making the problem a straightforward algebraic one. It enables us to substitute and isolate variables easily.

Factoring to Solve for Variables

After converting the system to exponential form, we often need to solve by isolating variables. This can involve factoring, a critical algebraic skill.

Once we substituted \(x = y^3\) into the equation \(4x = y^5\), we formed the equation:

• \(y^5 = 4y^3\)

This can be simplified to \(y^5 - 4y^3 = 0\), or \(y^3(y^2 - 4) = 0\) by factoring out \(y^3\). From here:

• The equation \(y^3 = 0\) gives \(y = 0\), but this is invalid because the base of a logarithm cannot be zero.
• The equation \(y^2 - 4 = 0\) simplifies to \(y^2 = 4\), leading us to \(y = ±2\).
• \(y = -2\) is invalid due to log base restrictions, leaving \(y = 2\) as the solution.

Factoring is crucial as it helps identify viable solutions while discarding those that don't fit the context.

The Role of Logarithms

Logarithms are the inverse operations of exponentiation. Understanding them is essential to solve equations like the ones in this exercise.

A logarithm \(\log_b a = c\) signifies that \(b\) raised to the power \(c\) equals \(a\), which is helpful in converting logarithmic expressions to exponential form.

• When solving logarithmic systems, ensure all base values are positive and not equal to one, as per the properties of logarithms.
• Check that the results adhere to these rules, especially after converting back and addressing errors involving invalid bases like negative numbers or zero.

Logarithms simplify complex multiplicative problems into additive ones, aiding significantly in solving equations and systems like those in this problem.