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Chapter 12: Problem 53

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{c} \ln x=4 \ln y \\ \log _{3} x=2+2 \log _{3} y \end{array}\right. $$

Short Answer

Expert verified
The solution is \( x = 81 \) and \( y = 3 \) or \( y = -3 \).

Step by step solution

01

Express in exponential form

Rewrite the given logarithmic equations in exponential form. Given equations are: 1) \( \ln x = 4 \ln y \)2) \( \log_{3} x = 2 + 2 \log_{3} y \).For equation 1: \( \ln x = 4 \ln y \) which can be rewritten as \( x = y^4 \).For equation 2: Rewrite it in exponential form: \( \log_{3} x - 2 \log_{3} y = 2 \Rightarrow \log_{3} \left( \frac{x} {y^2} \right) = 2 \). This implies \( \frac{x}{y^2} = 3^2 = 9 \).
02

Substitute and solve

Use the result from equation 1: \( x = y^4 \) in equation 2: \( \frac{y^4}{y^2} = 9 \) which simplifies to \( y^2 = 9 \).Solving for y, we get: \( y = 3 \) or \( y = -3 \).
03

Solve for x

Since \( y = 3 \) or \( y = -3 \), substitute these values back into \( x = y^4 \):If \( y = 3 \), then \( x = 3^4 = 81 \).If \( y = -3 \, then \ x = (-3)^4 = 81 \).Hence, for both \( y = 3 \) and \( y = -3 \), \( x = 81 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

logarithmic equations
Logarithmic equations involve variables inside logarithms. In this exercise, we have two logarithmic equations given:
1) \( \ln x = 4 \ln y \)
2) \( \log_{3} x = 2 + 2 \log_{3} y \).
To solve these equations, we need to rewrite them in simpler forms or exponential forms. This step makes it easier to manage and solve for the variables. Platforms like ours help students understand these processes clearly.
exponential form
Rewriting logarithmic equations in exponential form is a crucial step. Converting logs into exponentials simplifies complex log relationships.
Consider the equation \( \ln x = 4 \ln y \):
By using the property of logarithms, we transform it into \( x = y^4 \).

Similarly for \( \log_3 x = 2 + 2 \log_3 y \) we rearrange to get \( \log_3 \left( \frac{x}{y^2} \right) = 2 \) and then \( \frac{x}{y^2} = 3^2 = 9 \).
These exponential forms help us in the next steps of solving the equations.
substitution method
The substitution method makes solving systems of equations more manageable. We take one equation and solve it for one variable, then substitute into the other equation.
For this system, set \( x = y^4 \) from the first equation and replace x in the second equation:
\( \frac{y^4}{y^2} = 9 \).
It simplifies to \( y^2 = 9 \), giving \( y = 3 \) or \( y = -3 \).
By substituting these back, we simplify the system step-by-step, making the problem clearer and easier to work with.
solving systems of equations
Solving systems of equations means finding values that satisfy both equations simultaneously.
First, express in exponential forms:
From \( \ln x = 4 \ln y \): \( x = y^4 \)
From \( \log_3 x = 2 + 2 \log_3 y \): \( \frac{x}{y^2} = 9 \).
Next, use substitution with \( x = y^4 \). Solving for one system: \( y^2 = 9 \), so \( y = 3 \) or \( y = -3 \).
Finally, substitute these values to find \( x \):
If \( y = 3 \), then \( x = 81 \).
If \( y = -3 \), then \( x = 81 \) too.
Thus, the system has solutions \((x, y) = (81, 3) \) and \( (81, -3)\).

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