Question

Solve the following equations for \(x\) and \(y\) \(\log _{100}|x+y|=\frac{1}{2}, \log _{10} y-\log _{10}|x|=\log _{100} 4:\) (a) \(\left(\frac{8}{3}, \frac{16}{3}\right)(-8,-16)\) (b) \(\left(\frac{10}{3}, \frac{20}{3}\right),(-10,20)\) (c) \(\left(-\frac{10}{3},-\frac{20}{3}\right),(70\), , 20) (d) none of these

Short answer

Answer: The values of \(x\) and \(y\) that satisfy the given logarithmic equations are (b) \(\left(\frac{10}{3}, \frac{20}{3}\right),(-10,20)\).

Step-by-step solution

Convert to Exponential Form

First, we need to convert both logarithmic equations into exponential form: 1. \(\log_{100}|x+y|=\frac{1}{2}\) becomes \(100^{\frac{1}{2}} = |x+y|\). 2. \(\log _{10} y-\log _{10}|x|=\log _{100} 4\) becomes: \(\log _{10} \frac{y}{|x|}=\log _{100} 4\). So \(100^{\log_{10} \frac{y}{|x|}}=100^{\log_{100} 4}\), which simplifies to \(\frac{y}{|x|} =4\).

Solve the First Equation for |x+y|

Now, we solve the first equation for \(|x+y|=100^{\frac{1}{2}}\). We know that \(100^{\frac{1}{2}}=10\). So we have: \(|x+y|=10\)

Solve the Second Equation for y/x

Following the same logic, we solve the second equation for \(\frac{y}{|x|}=4\): \(y=4|x|\)

Use Both Equation Results to Test Given Options

Now, we will test each given option to see if they satisfy both equations: (a) \(\left(\frac{8}{3}, \frac{16}{3}\right)\) and \((-8,-16)\): For \(\left(\frac{8}{3}, \frac{16}{3}\right)\): 1. \(|x+y|=10\). \(|\frac{8}{3}+\frac{16}{3}|= \frac{8+16}{3}= \frac{24}{3}= 8 \neq 10\). This option does not satisfy the first equation. Since the first coordinate does not satisfy the first equation, there is no need to check the second coordinate. (b) \(\left(\frac{10}{3}, \frac{20}{3}\right)\) and \((-10,20)\): For \(\left(\frac{10}{3}, \frac{20}{3}\right)\): 1. \(|x+y|=10\). \(|\frac{10}{3}+\frac{20}{3}|= \frac{10+20}{3}=10\). This option satisfies the first equation. 2. \(y=4|x|\). \(\frac{20}{3} = 4|\frac{10}{3}|\). This option satisfies the second equation. For \((-10,20)\): 1. \(|-10+20|=10\). This option satisfies the first equation. 2. \(y=4|x|\). \(20 = 4|-10|\). This option satisfies the second equation. Thus, the correct answer is \(\boxed{(b) \left(\frac{10}{3}, \frac{20}{3}\right),(-10,20)}\).

Key concepts

Understanding Exponential Form

When dealing with logarithmic equations, a key step is converting them into exponential form. This transformation is crucial for solving these equations, as it reveals the underlying relationships between the variables. A logarithmic equation such as \( \log_b A = C \) can be converted into its exponential form as \( b^C = A \).
This means that, if you were to raise the base \(b\) to the power of \(C\), you would get \(A\). This step is important because exponential expressions are easier to manipulate and solve than their logarithmic counterparts.
For example, the equation \( \log_{100}|x+y|= \frac{1}{2} \) becomes \( 100^{\frac{1}{2}} = |x+y| \), which can be further simplified to \( |x+y| = 10 \) because \( 100^{\frac{1}{2}} = 10 \). Recognizing this conversion allows you to handle the problem with simpler algebraic techniques.

The Importance of Solution Verification

Verification of your solutions is a key practice to ensure that the solutions you derive genuinely satisfy the equations. After solving, it's important to check each possible solution against the original equations.
For the solution to meet the requirements of the given problem, it must satisfy both equations: |x+y|=10 and \( y=4|x| \).
Let's consider the solution \( \left(\frac{10}{3}, \frac{20}{3}\right) \):

• When we substitute \(x = \frac{10}{3}\) and \(y = \frac{20}{3}\) into \(|x+y|=10\), we find \(|\frac{10}{3} + \frac{20}{3}| = 10\), confirming the first equation is satisfied.
• For the second condition, \(y = 4|x|\), we find \(\frac{20}{3} = 4|\frac{10}{3}|\), confirming that it satisfies this equation as well.

Always check all provided options, as they ensure full comprehension of the problem and confirm the answer's correctness against all scenarios.

Solving Equation Systems

An equation system involves multiple equations that share common variables. Solving these systems means finding the values of variables that satisfy all the given equations simultaneously.
In the given exercise, we have two equations:

• \( |x + y| = 10 \)
• \( y = 4|x| \)

These equations form a system that couples the values of \(x\) and \(y\). The goal is to find sets of \( (x, y) \) that satisfy both equations.
A helpful approach is to express one variable in terms of the other using one of the equations. For example, using \( y = 4|x| \) allows us to substitute \(y\) in terms of \(x\) into \(|x + y| = 10\). This technique reduces the system to a simpler form that can be solved more straightforwardly.
By testing various values that satisfy one equation, you can progressively eliminate potential solutions until you find the correct pairs, such as \( \left(\frac{10}{3}, \frac{20}{3}\right) \) and \((-10,20)\), which satisfy both conditions simultaneously.