Problem 19 - Algebra 2

Question

Solve each equation. Check your answers. $$ 1.1+\ln x^{2}=6 $$

Step-by-Step Solution

Answer

The solution to the given equation \(1.1 + \ln x^{2} = 6\) is \(x = +\sqrt{e^{4.9}}\).

Step 1 Isolate the logarithmic term

Firstly, subtract 1.1 from both sides of the equation to isolate the logarithmic term: \(\ln x^{2} = 6 - 1.1 = 4.9\)

Step 2 Use the definition of natural logarithm

The natural logarithm can be written as exponentiation, i.e., \(e^{\ln a} = a\). Thus, we can re-write the equation as \(x^{2} = e^{4.9}\)

Step 3 Solve for x

To solve for x, take the square root of both sides of the equation: \(x = ±\sqrt{e^{4.9}}\). But, remember that we must check this solution, as \(\ln x^{2}\) is only defined for \(x ≠ 0\). This means that only the positive root is valid in this case.

Step 4 Check the solution

Substitute \(x\) into the original equation to check if it satisfies the equation. If we substitute \(x = +\sqrt{e^{4.9}}\), we get \(1.1 + \ln (\sqrt{e^{4.9}})^{2} = 1.1 + \ln e^{4.9} = 1.1 + 4.9 = 6\), confirming the solution is correct.

Key Concepts

Natural Logarithms

Natural logarithms use the base "e", which is an irrational constant approximately equal to 2.71828. In mathematics, natural logarithms are denoted as "\( \ln \)" and are used to solve problems related to growth and decay. They are the inverses of exponentiation with base \( e \).

When working with equations such as \( \ln x^2 = 4.9 \), the aim is to convert the logarithmic form into an exponential form to simplify the equation. This involves understanding that \( \ln a = b \) is equivalent to \( e^b = a \). Thus, \( \ln x^2 = 4.9 \) translates into \( x^2 = e^{4.9} \).

Using natural logarithms is particularly beneficial in exponential growth models, such as in calculating compound interest or understanding natural phenomena. Their applications extend into various fields, making them a vital part of mathematical literacy.

Exponentiation

Exponentiation is a mathematical operation involving a base and an exponent. It's the process of multiplying the base by itself a specified number of times, dictated by the exponent. For natural logarithms, exponentiation is used to "undo" the \( \ln \) operation, as \( e^{\ln a} = a \).

In our context, the equation \( x^2 = e^{4.9} \) is solved using the principle of exponentiation, recognizing that taking the square root of both sides will solve for \( x \).

When the equation is in the form of \( x^2 = e^{4.9} \), solving it requires finding \( x = \pm \sqrt{e^{4.9}} \). However, because we started with \( \ln x^2 \), we discard the negative root since the logarithm of a negative number is undefined.

This highlights the key aspect of exponentiation: understanding that it reverses logarithmic operations and that it must be handled with awareness of the properties of numbers involved.

Checking Solutions

Checking solutions is a crucial step in solving logarithmic equations to ensure accuracy and validity. This involves substituting the obtained solution back into the original equation to verify correctness.

In our exercise, after obtaining \( x = \sqrt{e^{4.9}} \), we substitute back to check: \( 1.1 + \ln (\sqrt{e^{4.9}})^2 = 1.1 + \ln e^{4.9} = 1.1 + 4.9 = 6 \). This confirms that the solution satisfies the original equation.

The checking process helps in mitigating simple computational errors that might have occurred during solving. It demonstrates the importance of verifying solutions, especially in cases involving complex functions like logarithms.

Include this step regularly to strengthen problem-solving skills, ensuring that the final answers are always trustworthy.