Question

Solve each equation. Express irrational solutions in exact form. $$ \ln x^{2}=(\ln x)^{2} $$

Short answer

The solutions are \( x = 1 \) and \( x = e^{2} \).

Step-by-step solution

Understand the given equation

The given equation is \(\ln x^{2}=(\ln x)^{2}\). Recognize that this is a logarithmic equation involving the natural logarithm function \(\ln\).

Apply the power rule for logarithms

Use the property of logarithms that states \(\ln(a^b) = b\ln(a)\). Thus, \(\ln x^{2}=2\ln x\). Now the equation becomes: \(2 \ln x = (\ln x)^{2}\).

Rearrange the equation

Rewrite the equation to facilitate solving it. We get \((\ln x)^{2} - 2 \ln x = 0\).

Factor the quadratic equation

Factor out \(\ln x\) from the equation: \(\ln x ((\ln x) - 2) = 0\).

Solve each factor

Set each factor equal to zero and solve: \(\ln x = 0\) gives \( x = e^{0} = 1 \). \(\ln x - 2 = 0\) gives \( \ln x = 2 \), hence \( x = e^{2} \).

Validate the solutions

Check that the solutions \(x = 1\) and \(x = e^{2}\) satisfy the original equation \(\ln x^{2}=(\ln x)^{2}\). Both values satisfy the equation.

Key concepts

natural logarithm

The natural logarithm, denoted as \(\backslash\text{ln}\), is a specific type of logarithm that uses the base \(e\). The constant \(e\) is approximately 2.71828 and is an important mathematical constant. Natural logarithms are used frequently in mathematics and science because of their natural properties when dealing with growth and decay processes. For example, when solving equations that involve exponential growth or when simplifying expressions with exponents.

Key points to remember about the natural logarithm:

• \( \backslashln 1 = 0 \) because \( e^0 = 1 \)
• \( \backslashln e = 1 \) because \( e^1 = e \)
• Its inverse function is the exponential function, \(e^x \)

Natural logarithms simplify solving equations involving exponentials, making them essential in calculus and algebra.

power rule for logarithms

The power rule for logarithms is a fundamental property that allows simplifying the logarithm of a number raised to an exponent. The rule states: \( \backslashln(a^b) = b \backslashln(a) \). This means that the logarithm of a power is the exponent multiplied by the logarithm of the base.

In our exercise, we used the power rule to transform the equation:
From: \( \backslashln x^2 = ( \backslashln x )^2 \)
To: \( 2 \backslashln x = ( \backslashln x )^2 \)
This transformation makes it easier to solve the equation by aligning it into a quadratic form.

The power rule can be very useful in a variety of problems involving logarithmic and exponential relationships, especially when combined with other logarithmic properties such as the product and quotient rules.

factoring quadratic equations

Factoring quadratic equations involves rewriting the equation as a product of its linear factors. This is feasible when the quadratic equation is set to zero. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \).

When factoring, you look for factors of \( a \times c \) that add up to \( b \).

In our case, the quadratic equation is in the form of logarithms: \( ( \backslashln x )^2 - 2 \backslashln x = 0 \),
We factored it as:
\( \backslashln x ( \backslashln x - 2 ) = 0 \)
This gave us two separate factors to solve:

• \( \backslashln x = 0 \) which results in \( x = e^0 = 1 \)
• \( \backslashln x - 2 = 0 \) which results in \( x = e^2 \)

Factoring is a key technique in algebra and is particularly useful in solving polynomial equations by breaking them down into more manageable pieces.

solving logarithmic equations

Solving logarithmic equations involves isolating the logarithmic term and then converting the equation into an exponential form to find the variable. It often includes utilizing logarithmic properties such as the power rule or change of base formula to simplify the equation.

Here are the steps we followed to solve the equation \( \backslashln x^2 = (\backslashln x)^2 \):

• Applied the power rule for logarithms: \( \backslashln x^2 = 2 \backslashln x \)
• Rearranged the equation: \( (\backslashln x)^2 - 2 \backslashln x = 0 \)
• Factored the quadratic equation: \( \backslashln x ( \backslashln x - 2 ) = 0 \)
• Solved each factor:
• \( \backslashln x = 0 \) gives \( x = 1 \)
• \( \backslashln x = 2 \) gives \( x = e^2 \)

It's important to verify the solutions within the original equation to ensure they are correct. Both solutions, \( x = 1 \) and \( x = e^2 \) satisfy the original equation, demonstrating the validity of our solved values. Solving logarithmic equations often requires careful manipulation of properties and thorough verification of solutions.