Chapter 5: Problem 64
Solve the systems $$\begin{aligned} &\log _{y} x=3\\\ &\log _{y}(4 x)=5 \end{aligned}$$
Short Answer
Expert verified
The solutions of the system are: \(x=8\) and \(y=2\).
Step by step solution
01
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: \(y^5 = 4x\).
02
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 \(2\).
03
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 \(x = 8\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
Here, we have:
- \(\log_{y} x = 3\)
- \(\log_{y}(4x) = 5\)
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:
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 \(4x = y^5\).
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:
Once we substituted \(x = y^3\) into the equation \(4x = y^5\), we formed the equation:
- \(y^5 = 4y^3\)
- 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.
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.
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.