Problem 63 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ 6+2 \ln x=5 $$

Step-by-Step Solution

Answer

The exact solution of the equation is \(x = e^{-1/2}\).

Step 1: Isolate the logarithmic term

First, we need to isolate the logarithmic term. This can be done by subtracting 6 from both sides of the equation. Doing so we get, 2 \ln x = 5 - 6 which simplifies to 2 \ln x = -1

Step 2: Solve for the variable in the logarithm

Next, we divide both sides of the equation by 2 to solve for \ln x. So, we get \ln x = -1/2. Recall the property of logarithms that states \(b = \ln a\) is equivalent to \(e^b = a\). So, here \(e^{-1/2} = x\)

Step 3: Check if the derived value is in the domain of the original logarithm

In this case, \(x = e^{-1/2}\) which is a positive number. Since logarithms are defined for positive numbers, the derived value for x is in the domain of the original logarithm. Hence, \(x = e^{-1/2}\) is the exact answer. And to get the decimal approximation of the value of x, use a calculator.

Key Concepts

Domain of Logarithmic Functions

The domain of a logarithmic function is crucial to understand when solving logarithmic equations. A logarithm is defined only for positive numbers. This means that any value inside the logarithm must be greater than zero for the expression to be valid. In the exercise provided, we have the equation with a logarithmic term, so we need to ensure that the solution we find is within the acceptable range for the logarithm.

In simple terms, the domain of any logarithmic function, such as \( \ln(x) \), is all positive real numbers. If our solution leads to \(x\) being negative or zero, it must be rejected. This step is crucial to ensure that our solution is mathematically sound and adheres to the properties of logarithmic functions.

Properties of Logarithms

Understanding the properties of logarithms helps in manipulating and solving logarithmic equations effectively. Here are some key properties that are often used:

• Product Rule: \(\log_b(MN) = \log_b(M) + \log_b(N)\)
• Quotient Rule: \(\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)\)
• Power Rule: \(\log_b(M^p) = p\cdot\log_b(M)\)
• Change of Base Formula: \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\)

In the specific equation provided \(2 \ln x = -1\), we used the power rule to simplify. Dividing both sides by 2 gives \( \ln x = -\frac{1}{2} \), and using the equivalence \(e^b = a\), we can convert the logarithmic equation into an exponential form. This technique helps in solving the equation easily and finding the value of \(x\) accurately.

Exact and Decimal Solutions

In solving logarithmic equations, finding the exact and decimal solutions is part of the process. First, solving an equation like \( \ln x = -\frac{1}{2} \) provides us with the exact solution for \(x\). Here, converting the logarithmic form to an exponential form gives us \(x = e^{-1/2}\). This expression represents the exact solution.

To obtain a decimal approximation of this value, we use a calculator. The base of the natural logarithm, \(e\), is approximately 2.71828. Applying this, we calculate the decimal value of \(e^{-1/2}\), which approximates to around 0.61 when rounded to two decimal places. Having both exact and decimal solutions is beneficial, especially in applications requiring precise measurements or when the context asks for a specific level of precision.