91Ó°ÊÓ

Linear equations are fundamental in algebra and involve variables raised to the first power. They look like straight lines when graphed. In simple terms, a linear equation has the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable.

To solve linear equations:

• Isolate the variable on one side of the equation by using inverse operations, like addition, subtraction, multiplication, or division.
• Ensure that the variable's coefficient is one, if possible, to simplify solving.

In the exercise, the equation \((x-1)^{\log (x-1)} = 100(x-1)\) was manipulated using substitution with \(y = x-1\), which reduced it to a form involving linear interpretation \(y^{\log y - 1} = 100\). Simplifying gives the value as \(y = 10\), helping to solve for the original variable \(x = 11\).
Understanding substitution in equations often brings them down to a recognizable linear form, making them easier to solve.

Quadratic equations form the next step in algebra progression, dealing with polynomials of degree 2. A standard quadratic equation looks like \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable.

Key methods to solve quadratic equations include:

• Factoring the expression into products of simpler binomials.
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Completing the square to transform the equation into the form \((x-h)^2 = k\).

For the equation \(4^{x} - 2^{x+1} = 3\) in the exercise, we substituted with \(y = 2^x\). This converts the expression into a quadratic form \(y^2 - 2y - 3 = 0\), which can be factored into \((y - 3)(y + 1) = 0\). The solution \(y = 3\) leads to \(x = \log_2 3\). This transformation highlights the importance of recognizing hidden quadratic forms in equations.

Logarithmic equations involve logarithms of expressions set equal to a number. The basic form is \(\log_b(x) = y\), where \(b\) is the base, \(x\) is the argument, and \(y\) is the result.

To solve equations involving logarithms, consider:

• Using the properties of logarithms, like \(\log_b(xy) = \log_b x + \log_b y\) or \(\log_b(x/y) = \log_b x - \log_b y\).
• Converting logarithms with different bases using the change of base formula: \(\log_b a = \frac{\log_c a}{\log_c b}\).
• Exponentiating to simplify: If \(\log_b(x) = y\), then \(x = b^y\).

In part \(b\) of the exercise, the equation \(\log_{2} x + \log_{4} x + \log_{8} x = 11\) was approached using the change of base formula, transforming it into common logarithms. After simplification, the expression was solvable, leading to \(x = 8\). Grasping these techniques broadens understanding of how different forms of logarithmic equations can appear and be solved.