Problem 65 - Essential Calculus Early Transcendentals

Question

Solve each equation for \(x\) $$ \text { (a) }2^{x-5}=3 \quad \text { (b) } \ln x+\ln (x-1)=1 $$

Step-by-Step Solution

Answer

a) \(x \approx 6.585\); b) \(x \approx 2.193\).

Step 1: Isolate the exponent in equation (a)

For equation (a) \(2^{x-5} = 3\), isolate \(x\) by taking the log base 2 of both sides. This gives you \(x-5 = \log_2(3)\).

Step 2: Solve for the variable in equation (a)

To solve for \(x\) in equation (a), add 5 to both sides of the equation: \(x = \log_2(3) + 5\). Calculate \(\log_2(3)\) using the change of base formula, which yields approximately 1.585. So, \(x \approx 6.585\).

Step 3: Use logarithm properties for equation (b)

For equation (b) \(\ln x + \ln(x-1) = 1\), use the property \(\ln a + \ln b = \ln(ab)\) to combine the logs. This gives you \(\ln(x(x-1)) = 1\).

Step 4: Exponentiate to remove the natural log in equation (b)

Exponentiate both sides of the equation from Step 3 to eliminate the natural logarithm: \(x(x-1) = e^1\). This gives you the quadratic equation \(x^2 - x = e\).

Step 5: Solve the quadratic equation for equation (b)

Rearrange the quadratic equation to \(x^2 - x - e = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1, b=-1, c=-e\). Solve for \(x\) to get the solutions: \(x = \frac{1 \pm \sqrt{1 + 4e}}{2}\). Calculate the approximate values using \(e \approx 2.718\), which results in \(x \approx 2.193\) (only positive solution).

Key Concepts

Change of Base Formula

The Change of Base Formula is a crucial tool when working with logarithms, especially when you need to evaluate a log with a base that your calculator does not support directly. Let's explore it in a simple way.

When you have a logarithm with a base that isn't feasible to calculate directly, such as \ \( \log_b(a) \ \), you can convert it using a different base like 10 or \(e\) (natural logarithm base). The formula looks like this:

\[\log_b(a) = \frac{\log_k(a)}{\log_k(b)}\]

This means you can choose any log base \(k\) that is convenient, often \(10\) or \(e\). For example, in the exercise, when finding \(\log_2(3)\), you apply the change of base formula. Using base 10:

• \(\log_2(3) = \frac{\log_{10}(3)}{\log_{10}(2)}\)

This conversion allows you to approximate \(\log_2(3) \approx 1.585\), enabling you to solve the equation \(2^{x-5} = 3\) effectively.

Natural Logarithm Properties

Natural logarithms, denoted by \(\ln\), have unique properties that are very useful in simplifying and solving equations. The natural logarithm uses the constant \(e\) as its base, where \(e \approx 2.718\). One of the most handy properties of logarithms is the product rule: \(\ln(a) + \ln(b) = \ln(ab)\).

In the original problem, equation \((b)\) makes use of this property to combine two logarithmic expressions:

• \(\ln(x) + \ln(x-1) = 1\)
• \(\Rightarrow \ln(x(x-1)) = 1\)

This simplification helps transform the logarithmic equation into an exponential form. Exponentiating both sides, you write:

\[x(x-1) = e^1\]

This allows the problem to transition smoothly into solving a quadratic equation, as you are essentially removing \(\ln\) by raising \(e\) to the power of both sides.

Quadratic Equations

Quadratic equations are polynomials of degree two, generally in the form \(ax^2 + bx + c = 0\). Solving these equations is fundamental in algebra and can be done using several methods, such as factoring, completing the square, or the famous quadratic formula.

In the exercise above, equation \((b)\) ultimately reduced to a quadratic equation after removing the natural logarithm using exponentiation:

• \(x^2 - x = e\)
• \(\Rightarrow x^2 - x - e = 0\)

The solution involves the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

For our specific case, the coefficients are \(a=1\), \(b=-1\), and \(c=-e\). Plugging these into the formula provides the solution for \(x\). Calculating gives us:

• \(x = \frac{1 \pm \sqrt{1 + 4e}}{2}\)
• Since \(e \approx 2.718\), we only consider the positive root \(x \approx 2.193\), since \(x\) must be greater than zero.

This illustrates how foundational algebraic skills are often reinforced through solving equations that arise from natural logarithms and can lead to quadratic equations.