91Ó°ÊÓ

A quadratic equation is a type of polynomial equation characterized by its highest degree, which is two. It generally has the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). These equations can typically be solved using methods such as factoring, using the quadratic formula, or completing the square.

In our exercise, we solved for \( x \) in the equation derived from logarithms: \( x^2 - 3x = 0 \). By factoring, we identified the solutions of the quadratic equation. The expression factored gives us \( x(x - 3) = 0 \).

• Setting each factor equal to zero, we find the potential solutions: \( x = 0 \) and \( x = 3 \).
• However, considering the domain from the integral bounds, \( x \) must be greater than 1, which leads us to select \( x = 3 \) as the valid solution.

Quadratic equations are crucial because they appear frequently in various branches of mathematics and real-world scenarios, such as physics and engineering.

Logarithms are a mathematical concept that serve as the inverse of exponentiation. The logarithm of a number is essentially the power to which the base, typically \( 10 \) or \( e \), must be raised to produce that number.

In the step-by-step solution of our exercise, the integral transforms resulted in logarithmic expressions. We employed the property that the integral of \( \frac{1}{t} \) over a certain range is the natural log of the endpoints. Specifically, \( \int \frac{1}{t} \, dt = \ln |t| \), leveraging this to evaluate definite integrals to logarithms of \( x \) and constants.

We used these log properties:

• \( \ln(ab) = \ln(a) + \ln(b) \), allowing us to combine logarithms.
• \( a \ln(b) = \ln(b^a) \), useful in simplifying terms in the equation.

Moreover, to solve the equation \( \ln(3x) = \ln(x^2) \), it was critical to understand that if \( \ln(a) = \ln(b) \), then necessarily \( a = b \), leading us naturally to the quadratic equation step.

Definite integrals compute the accumulation of quantities, such as area under a curve, between given limits \( a \) and \( b \). In our problem, the integrals of the function \( \frac{1}{t} \) from \( \frac{1}{3} \) to \( x \) and from \( 1 \) to \( x \) were evaluated to determine a balance of areas represented algebraically.

The integral \( \int_a^b \frac{1}{t} \, dt \) gives us a result related to the difference in natural logarithms of the bounds. This played out in the steps:

• The calculation for the interval \( \left[\frac{1}{3}, x\right] \) resulted in \( \ln(3x) \), capturing the change across that range.
• For the \( [1, x] \) range, it simplified to \( \ln(x) \), reflecting a simpler change.

Definite integrals are fundamental in calculus for analyzing continuous change and finding exact values for accumulated quantities. In our problem-solving process, they provided the essence of the initial equation to be manipulated and eventually solved.