Question

If \(p \in R\) and \(q=\log _{x}\left(p+\sqrt{p^{2}+1}\right)\), then the value of \(p\) interms of \(x\) and \(q\) is (1) \(\frac{x^{q}+x^{-q}}{2}\) (2) \(\frac{x^{4}-x^{-q}}{2}\) (3) \(\mathrm{x}^{\mathrm{q}}+\mathrm{x}^{-9}\) (4) \(\mathrm{x}^{9}-\mathrm{x}^{-4}\)

Short answer

Question: Determine the value of \(p\) given that \(p\) is an element of the set \(R\) and \(q = \log_x(p+\sqrt{p^2+1})\), where \(x\) is a real number. Answer: \(p = \frac{x^{q}+x^{-q}}{2}\)

Step-by-step solution

Express \(q\) as an equation in terms of \(p\) and \(x\)

We are given \(q = \log_x(p+\sqrt{p^2+1})\). Using the logarithm property that states \(y = \log_b{x}\) is equivalent to \(b^y = x\), our equation can be rewritten as: \(x^q = p+\sqrt{p^2+1}\)

Solve the equation for \(p\)

Subtract \(p\) from both sides of the equation: \(x^q - p = \sqrt{p^2+1}\) Now, square both sides to get rid of the square root: \((x^q - p)^2 = p^2 + 1\) Expand the square on the left side and simplify: \((x^{2q} - 2x^qp+p^2) = p^2 + 1\) Subtract \(p^2\) from both sides: \(x^{2q}-2x^qp = 1\) Now, we'll solve for \(p\): \(p(2x^q) = x^{2q}-1\) \(p = \frac{x^{2q}-1}{2x^q}\)

Compare the result to the given options

We have found that \(p = \frac{x^{2q}-1}{2x^q}\). By observing the given options, we can tell that we need to rewrite this expression using properties of exponents: \(p = \frac{x^q( x^q - x^{-q})}{2}\) \(p = \frac{x^q + x^{-q}}{2}\) Comparing this result to the given options, we find that it matches option (1): \(p = \frac{x^{q}+x^{-q}}{2}\)

Key concepts

Logarithm Properties

Understanding logarithm properties is vital for solving logarithmic equations you'll come across in mathematics. A logarithm answers the question: To what power must we raise a certain base to obtain a given number? The two most fundamental properties that one should be aware of are the Product Property and the Quotient Property.

The Product Property states that \( \log_b(mn) = \log_b(m) + \log_b(n) \), meaning the log of a product is equal to the sum of the logs. Similarly, the Quotient Property indicates that \( \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \). These properties can be used to break down complex logarithmic problems into more manageable pieces.

Additionally, the Power Property \( \log_b(m^n) = n \cdot \log_b(m) \) suggests that a log raised to an exponent allows the exponent to be pulled out in front as a coefficient. This property is particularly useful when dealing with equations where the variable is an exponent within a logarithm. By grasping these properties, you can simplify and solve complex logarithmic equations, making it an essential toolkit in your math arsenal.

Solving Logarithmic Equations

Solving logarithmic equations may seem daunting at first, but with a clear understanding of logarithm properties, it becomes much simpler. Here's a strategy to tackle these problems: First, if needed, use the properties of logarithms to combine or expand logs, aiming to isolate the logarithmic terms.

After isolating, convert the logarithmic equation into exponential form, which is the very crux of solving most logarithmic equations. If we consider an equation of the form \( \log_b(x) = y \), then the equivalent exponential form is \( b^y = x \). Employing this allows us to deal with more familiar, traditional algebraic operations.

Sometimes the solution requires further algebraic manipulation, such as squaring both sides to eliminate a square root, like in the exercise above. Always remember to check for extraneous solutions since not all results produced algebraically may be valid in the original logarithmic equation context. Solving logarithmic equations step-by-step in a methodical way is the key to mastering them.

Exponent Properties

Exponent properties, also known as the laws of exponents, are rules that govern arithmetic operations involving powers. They help us simplify expressions and solve equations where variables are raised to powers. The most often used properties are the Product of Powers, stating that \( b^m \cdot b^n = b^{m+n} \), and the Quotient of Powers, indicating that \( \frac{b^m}{b^n} = b^{m-n} \) if \( b \eq 0 \).

Another essential property is the Power of a Power, which states that \( (b^m)^n = b^{mn} \). The Power of a Product \( (mn)^p = m^p \cdot n^p \) and the Power of a Quotient \( (\frac{m}{n})^p = \frac{m^p}{n^p} \) round off the set of critical exponent properties to remember.

These rules simplify the process of operating with exponents and are particularly important when you evaluate and compare expressions, as seen in the step-by-step solution referencing the exercise above. By combining the knowledge of logarithm properties and exponent properties, we can solve a large variety of algebraic equations effectively and efficiently.