Problem 239 - Precalculus

Question

For the following exercises, use a graphing calculator to find approximate solutions to each equation. $$\log (x-1)+2=\ln (x-1)+2$$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 11 \).

Step 1: Simplify the Equation

First, notice that both sides of the equation have the same constant, "+2". Cancel out the "+2" from both sides to get:\[ \log(x-1) = \ln(x-1) \]

Step 2: Use Change of Base Formula

We want to work with the same type of logarithm. The change of base formula can be applied to the right side. Recall the change of base formula: \[ \log_b(a) = \frac{\ln(a)}{\ln(b)} \]Thus, \[ \log(x-1) = \frac{\ln(x-1)}{\ln(10)} \]

Step 3: Set Up the Equation for Solving

By substituting the base-change formula into the equation, you get:\[ \log(x-1) = \log(x-1)/\ln(10) \]After simplifying, this becomes:\[ 1 = \frac{1}{\ln(10)} \]

Step 4: Solve the Equation for x

Now, since the equation involves log of \(x-1\), it implies\[ x-1 = 10^1 \]Solve for \(x\):\[ x - 1 = 10 \]Thus:\[ x = 11 \]

Step 5 (Check the Solution): Ensure the Solution Makes Sense

Substitute \(x = 11\) back into the original equation to verify:\[ \log(11-1) + 2 = \ln(11-1) + 2 \]Which simplifies as both sides equal because both functions return the same value for \(x - 1 = 10\).

Key Concepts

logarithms

Logarithms are incredibly useful in mathematics and appear in a variety of real-world applications such as solving problems involving exponential growth or decay. A logarithm answers the question: "To what power must we raise a certain number, called the base, to obtain another number?"
The general formula for a logarithm can be written as:

• If \( b^y = x \), then \( \log_b(x) = y \)

Here, \( b \) represents the base of the logarithm, \( x \) is the number we want to find the logarithm for, and \( y \) is the exponent to which the base is raised.
Understanding logarithms is key in solving equations that involve exponential terms, as they help in rewriting the equations in a more manageable form.

change of base formula

The change of base formula allows you to rewrite a logarithm in terms of logarithms with different bases, which can be particularly helpful when solving equations. This comes in handy when you are using calculators, as most calculators are equipped to handle base-10 logarithms (common logarithms) and natural logarithms (base \( e \)).
The formula is written as:

• \( \log_b(a) = \frac{\ln(a)}{\ln(b)} \)

Where \( a \) is the argument of the logarithm, and \( b \) is the base you are converting from.
Using the change of base formula allows you to convert any logarithm into a form that can be easily calculated using standard calculator functions. This conversion is a crucial step in simplifying and solving logarithmic equations, as seen when we equate \( \log(x-1) \) with \( \ln(x-1) \).

solving equations

Solving equations with logarithms often involves simplifying and using properties of logarithms to isolate the variable. In this case, both sides of the equation originally included a constant, which we removed for simplicity:

• \( \log(x-1) + 2 = \ln(x-1) + 2 \) reduced to \( \log(x-1) = \ln(x-1) \)

The aim here is to get the variable by itself.

• To do this, we applied the change of base formula for \( \ln(x-1) \) to express it in base-10 logarithms.
• Simplifying the equation further, we derived a condition where \( x - 1 \) must satisfy the equation: \( 1 = \frac{1}{\ln(10)} \).
• Through this simplification, we solved for \( x \), yielding \( x = 11 \).

Understanding these basic steps of isolating variables and making equations easier to handle is fundamental in algebra.

checking solutions

After solving an equation, it's important to verify that the solution is indeed correct by substituting it back into the original equation. This step ensures that there hasn't been any error in computation or incorrect application of the formulas.
In our exercise, we found \( x = 11 \) as the solution.

• We checked this by substituting \( 11 \) back into the simplified original equation:
• \( \log(11-1) + 2 = \ln(11-1) + 2 \)
• Both sides were equal after substitution.

Checking ensures the integrity of problem-solving processes and helps reinforce correct mathematical reasoning. In examinations or practical applications, verifying solutions can prevent errors that might otherwise go unnoticed. Remember, the solution makes sense if, after substituting, both sides of the equation are equal. This small step is crucial for precision in mathematics.