Problem 93
Question
$$ y=\ln \frac{1}{x+\sqrt{x^{2}-1}} $$
Step-by-Step Solution
Verified Answer
The simplified logarithmic function could either be \(y = -\ln(x+1)\) or \(y = -\ln(x-1)\), depending on the specific root chosen in step 2.
1Step 1: Apply properties of logarithms
The given function can be rewritten as \(y=\ln(1) - \ln(x+\sqrt{x^{2}-1})\). The logarithm of 1 in any base equals to 0, therefore, \(y= -\ln(x+\sqrt{x^{2}-1})\)
2Step 2: Apply substitution
Let \(t = x+\sqrt{x^{2}-1}\). Then, \(t-x = \sqrt{x^{2}-1}\). Square both sides, and get \(t^{2} - 2tx + x^{2} = x^{2} - 1\). If we simplify, we get \(t^{2} - 2tx + 1 = 0\). It is known that any quadratic equation of the form \(ax^{2} + bx + c = 0\) has its roots given by \(-b \pm \sqrt{b^{2} - 4ac}/2a \). Using that formula, the roots will be \(t = x \pm 1\)
3Step 3: Express original function in terms of new variable
Both roots can be substituted back into the function formed in step 2: \(y = -\ln(x+1)\) or \(y = -\ln(x-1)\)
Key Concepts
Properties of LogarithmsSubstitution Method in AlgebraSolving Quadratic Equations
Properties of Logarithms
Understanding how logarithms work is crucial in solving equations that include them. Logarithms, simply put, are another way to represent exponents. They have certain properties which can simplify complex expressions significantly. The most commonly used properties include the product rule, quotient rule, and the power rule.
The product rule of logarithms states that the log of a product is equal to the sum of the logs: \[\log_b(mn) = \log_b(m) + \log_b(n)\]. The quotient rule, as seen in the original exercise with \(y=\ln \frac{1}{x+\sqrt{x^2-1}}\), states that the log of a quotient is the difference between the logs: \[\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\]. This property allowed us to rewrite the given expression for easier manipulation.
Knowing these rules is integral to simplifying logarithmic equations and setting them up for further solving steps.
The product rule of logarithms states that the log of a product is equal to the sum of the logs: \[\log_b(mn) = \log_b(m) + \log_b(n)\]. The quotient rule, as seen in the original exercise with \(y=\ln \frac{1}{x+\sqrt{x^2-1}}\), states that the log of a quotient is the difference between the logs: \[\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\]. This property allowed us to rewrite the given expression for easier manipulation.
Knowing these rules is integral to simplifying logarithmic equations and setting them up for further solving steps.
Substitution Method in Algebra
The substitution method is a powerful tool in algebra that simplifies equations by reducing the number of variables or by transforming an equation into a more familiar form. It involves replacing one variable with another variable or expression that is equivalent. In the context of the given exercise, a substitution \(t = x + \sqrt{x^2 - 1}\) is made.
Upon squaring both sides and rearranging, we transformed a radical equation into a recognizable quadratic equation. Without this substitution, solving the original logarithmic equation directly would be very complicated. The substitution method can also be utilized to solve systems of equations, where one equation is solved for one variable and that expression is substituted into the other equation.
Upon squaring both sides and rearranging, we transformed a radical equation into a recognizable quadratic equation. Without this substitution, solving the original logarithmic equation directly would be very complicated. The substitution method can also be utilized to solve systems of equations, where one equation is solved for one variable and that expression is substituted into the other equation.
Solving Quadratic Equations
Quadratic equations are polynomials of the second degree, generally resembling the standard form \(ax^2 + bx + c = 0\). There are several methods to solve these, such as factoring, completing the square, and the quadratic formula. The quadratic formula \(x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)\) is a universally applicable method that finds the roots of any quadratic equation.
In our exercise, after using substitution, we derived a quadratic in terms of \(t\): \(t^2 - 2tx + 1 = 0\). The quadratic formula provided us with the solutions \(t = x \pm 1\). Recognizing that we are working with a quadratic equation allows us to apply the quadratic formula and thus find the roots efficiently. This skills is pivotal in proceeding with solving a multitude of algebraic problems.
In our exercise, after using substitution, we derived a quadratic in terms of \(t\): \(t^2 - 2tx + 1 = 0\). The quadratic formula provided us with the solutions \(t = x \pm 1\). Recognizing that we are working with a quadratic equation allows us to apply the quadratic formula and thus find the roots efficiently. This skills is pivotal in proceeding with solving a multitude of algebraic problems.