91Ó°ÊÓ

Logarithms are a fundamental concept in mathematics used to solve equations involving exponential growth and decay. At its core, a logarithm answers the question: "To what power must a base be raised, to yield a certain number?" This is often expressed as

• \( \log_{b}(x) = y \)

This means that raising the base \( b \) to the power \( y \) gives \( x \). It is an inverse operation to exponentiation, similar to how subtraction is the inverse of addition.
When working with logarithms, keep in mind:

• They only apply to positive numbers.
• The base of a logarithm must always be positive and not equal to 1.

In real-world applications, logarithms help model very large or very small numbers, such as in finance or science. For example, the Richter scale for earthquakes is logarithmic.
In solving this exercise, the goal is to find all possible values of \( y \) that satisfy the given logarithmic equation.

The bridge from logarithms to solving equations often lies in converting the logarithmic equation into an exponential form. This means taking the understood logarithmic statement

• \( \log_{(y-1)} 4 = 2 \)

and rearranging it using the definition of a logarithm.
By following this process, we can see that it converts to:

• \((y-1)^2 = 4\)

Using the exponential form, we have transformed the problem from a logarithm into a quadratic expression, which many find more straightforward to solve.
Converting to exponential form is a powerful tool because it simplifies the equation, making it easier to see how we can solve for the variable involved, in this case \( y \).

• This allows us to use algebraic techniques, such as factoring or using the quadratic formula, to find the solutions.

With the equation in exponential form, solving for \( y \) becomes a more familiar task. In this exercise, our equation \((y-1)^2 = 4\) is a quadratic one. Quadratic equations can have two, one, or no solutions at all. Here are the steps to solve it:

• Recognize \((y-1)^2 = 4\) implies we need to think in terms of square roots: \(y-1 = \pm 2\).

This leads us to separate into two cases:

• Case 1: \(y-1 = 2\) which gives \(y = 3\).
• Case 2: \(y-1 = -2\) which gives \(y = -1\).

These solutions mean that \( y \) can take either value to satisfy the original logarithmic equation.
However, it's crucial to consider the domain of the original logarithm: the base \( y-1 \) must be positive and not equal to 1. After verifying, \( y = 3 \) is valid, but \( y = -1 \) is not since \( y-1 \) cannot be negative for a logarithm. Thus, our valid solution is \( y = 3 \).

• Always check that proposed solutions maintain all original constraints of the logarithmic expressions.