Problem 51 - Contemporary Precalculus

Question

Solve the equation. $$\ln \left(x^{2}+1\right)-\ln (x-1)=1+\ln (x+1)$$

Step-by-Step Solution

Answer

Question: Solve the equation \(\ln \left(x^{2}+1\right)-\ln (x-1)= 1+\ln (x+1)\) for x. Answer: \(x=\frac{-(2e-1) \pm \sqrt{(2e-1)^{2} - 4(1-e)(1-e)}}{2(1-e)}\)

Step 1: Combine the logarithms on the left-hand side

Using the properties of logarithms, we can rewrite \(\ln \left(x^{2}+1\right)-\ln (x-1)\) as $$\ln \frac{x^{2}+1}{x-1}$$ The equation now becomes $$\ln \frac{x^{2}+1}{x-1}= 1+\ln (x+1)$$

Step 2: Get rid of the logarithms

In order to get rid of the logarithms, we can use the property that if \(\ln a = \ln b\), then \(a = b\). Therefore, we can rewrite the equation as $$\frac{x^{2}+1}{x-1} = (x+1)e^1$$

Step 3: Simplify and create a quadratic equation

Multiply both sides by \((x-1)\) to get rid of the fraction and simplify: $$x^{2}+1 = (x+1)e(x-1)$$ Now, expand \((x+1)e\): $$x^{2}+1 = ex^{2} - e x + e x - e$$ Rearrange the terms to create a quadratic equation: $$(1-e)x^{2} + (2e-1)x + (1-e) = 0$$

Step 4: Solve the quadratic equation

To solve the quadratic equation, we will use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In our case, we have \(a = 1-e, b = 2e-1,\) and \(c = 1-e\). Substitute these values into the quadratic formula: $$x = \frac{-(2e-1) \pm \sqrt{(2e-1)^{2} - 4(1-e)(1-e)}}{2(1-e)}$$ At this point, we can use a calculator to numerically solve for the real values of x.

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is crucial in solving logarithmic equations. Logs are not just another function; they have unique properties that allow us to simplify complex expressions. Firstly, the logarithm of a product can be expressed as the sum of the logarithms of its factors. This is expressed as \[\log(a \times b) = \log(a) + \log(b)\]. Conversely, the logarithm of a quotient is the difference of the logarithms: \[\log\left(\frac{a}{b}\right) = \log(a) - \log(b)\]. Another important property is that the logarithm of an exponent allows you to bring the exponent down as a coefficient: \[\log(a^x) = x \cdot \log(a)\].
Using these properties, complex logarithmic expressions can often be rewritten in simpler terms that are easier to solve, as demonstrated in the step-by-step solution for the exercise, where we see the subtraction of two logarithms transformed into a single log of a division.

Solving Quadratic Equations

When you encounter a quadratic equation, where the highest power of the variable is a square, the solution may require several steps to simplify before applying any formulas. The typical form of a quadratic equation is \[ax^2 + bx + c = 0\], where 'a', 'b', and 'c' represent known numbers, and 'x' is what we're solving for.
There are several methods to solve for 'x', like factoring, completing the square, or using the quadratic formula, which is a reliable last resort when other methods fail. The quadratic formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], encompassing the ‘plus or minus’ expression, which indicates there may be two solutions. In the exercise above, we apply the quadratic formula after rearranging the expression into the standard form. This method provides a systematic approach to finding the 'x' values that satisfy the original equation.

Natural Logarithm

The natural logarithm, denoted as 'ln', is a specific type of logarithm with the number 'e' (approximately equal to 2.718281828459) as its base. 'e' is an irrational and transcendental number, fundamental to mathematics, much like pi (\[\pi\]). The natural logarithm of a number 'x' is the power to which 'e' must be raised to obtain that number.
The equation \[\ln(x) = y\] is equivalent to the exponential form \[e^y = x\]. In the context of the given exercise, when we have an equation that equates two logs, as seen after using the logarithm properties, we can remove the 'ln' notation by setting the arguments equal if the logarithms have the same base. This property enables us to form an equation like \[\frac{x^{2}+1}{x-1} = (x+1)e\], which after further manipulation leads to a quadratic equation and ultimately to finding the value of 'x'. Understanding the relationship between the natural logarithm and its base is essential when dealing with natural log equations.