Question

If \(\log _{(x-y)}(x+y)=5\), then what is the value of \(\log _{x^{2}-y^{2}}\left(x^{2}-2 x y+y^{2}\right) ?\) (1) 1 (2) \(\frac{\sqrt{5}}{3}\) (3) \(\frac{1}{3}\) (4) 0

Short answer

Answer: The correct value for \(\log_{x^{2}-y^{2}}\left(x^{2}-2xy+y^{2}\right)\) is -4, which doesn't match any of the provided options.

Step-by-step solution

Write down the given logarithm expression

Given, \(\log _{(x-y)}(x+y)=5\). Our goal is to find the value of \(\log_{x^{2} - y^{2}}\left( x^{2} - 2xy + y^{2}\right)\) using this information.

Convert the given logarithm expression into exponential form

Using the definition of logarithm, we can write the given expression as an exponential equation: \((x-y)^{5}=(x+y)\).

Manipulate the required expression using logarithm properties

We want to find the value of \(\log_{x^{2}-y^{2}}\left(x^{2}-2xy+y^{2}\right)\). Using the difference of square factors and logarithm properties, we have: \(\log_{x^{2}-y^{2}}\left(x^{2}-2xy+y^{2}\right) = \log_{(x+y)(x-y)}\left((x-y)(x-y)\right)\) Using logarithm property, \(\log_{a}\left(bc\right)=\log_{a}{b}+\log_{a}{c}\), we get: \(\log_{(x+y)(x-y)}\left((x-y)(x-y)\right) = \log_{(x+y)(x-y)}\left((x+y)(x-y)\right) - \log_{(x+y)(x-y)}(x-y)\) Now, notice that \(\log_{(x+y)(x-y)}\left((x+y)(x-y)\right) = 1\) and we have: \(\log_{x^{2}-y^{2}}\left(x^{2}-2xy+y^{2}\right) = 1 - \log_{(x+y)(x-y)}{(x-y)}\)

Use the given expression to substitute the value

We know that \(\log_{(x-y)}(x+y)=5\). Therefore, we can rewrite our above expression as: \(\log_{x^{2}-y^{2}}\left(x^{2}-2xy+y^{2}\right) = 1 - \frac{\log_{(x+y)(x-y)}{(x+y)}}{\log_{(x+y)(x-y)}{(x-y)}}\) Since \(\log_{(x-y)}(x+y) = 5\), we have \(\log_{(x+y)(x-y)}{(x+y)} = 5\log_{(x+y)(x-y)}{(x-y)}\), thus: \(\log_{x^{2}-y^{2}}\left(x^{2}-2xy+y^{2}\right) = 1 - \frac{5\log_{(x+y)(x-y)}{(x-y)}}{\log_{(x+y)(x-y)}{(x-y)}}\) Finally, the expression simplifies to: \(\log_{x^{2}-y^{2}}\left(x^{2}-2xy+y^{2}\right) = 1 - 5\)

Get the final answer

After simplifying the above expression, we get: \(\log_{x^{2}-y^{2}}\left(x^{2}-2xy+y^{2}\right) = -4\) None of the given options matches the result. So, the correct answer is not in the provided options.

Key concepts

Logarithmic Expressions

Understanding logarithmic expressions is essential when solving a variety of mathematical problems. A logarithm essentially tells us what exponent we need to raise a number, known as the base, to get another number. It's written in the form \( \log_b{a} \) where 'b' is the base, 'a' is the number we're trying to find the exponent for, and the answer is the power that 'b' is raised to, to get 'a'.

An example of a logarithmic expression is \( \log_2{8} = 3 \) because \( 2^3 = 8 \.\) In this case, '2' is the base, '8' is the number we want to get by raising the base to some power, and '3' is the power which makes this equation true. This type of expression lends itself to a variety of applications, including solving for unknown variables and changing the form of equations for further analysis.

In the given exercise, we're asked to deduce the value of a particular logarithmic expression involving variables. This requires understanding how to handle log functions and their interplay with algebraic expressions.

Exponential Form Conversion

When working with logarithms, it's often useful to convert logarithmic expressions to their exponential form for easier manipulation. This conversion leverages the basic definition of logarithms: if \( \log_b{a} = c \) then it follows that \( b^c = a \.\) Here, 'b' is raised to the power 'c' to get 'a'.

This process is especially handy when you're given a logarithmic equation and need to find the value of the variable. By converting to exponential form, you can use the rules of exponents to simplify and solve algebraic equations. For example, in the problem \( \log _{(x-y)}(x+y)=5 \) from the given exercise, the exponential form conversion allows us to interpret the equation as \( (x-y)^5 = x+y \.\) It is a crucial step that unlocks the capability to use the properties of powers and roots in further problem-solving steps. In essence, exponential form makes complex logarithmic problems more tangible and solvable.

Logarithm Properties

Logarithms come with a toolkit of properties that can significantly simplify solving problems. Among these properties are:

• The Product Rule: \( \log_b{(mn)} = \log_b{m} + \log_b{n} \) states that the log of a product is the sum of the logs.
• The Quotient Rule: \( \log_b{(\frac{m}{n})} = \log_b{m} - \log_b{n} \) tells us that the log of a quotient is the difference of the logs.
• The Power Rule: \( \log_b{(m^c)} = c \cdot \log_b{m} \) indicates that the log of a power is the exponent times the log of the base.
• The Change of Base Formula: This allows us to convert logs of one base into another, which is often helpful for calculation purposes.

These properties are not just mathematical curiosities; they are powerful tools that can be applied to break down and simplify logarithms involving complex expressions. In the exercise at hand, recognizing that the expression within the logarithm is a difference of squares and then applying the product rule lets us rewrite and simplify the logarithm efficiently. Mastery of these properties means turning what could be a daunting logarithmic expression into a manageable and solvable equation.