Problem 57 - Elementary and Intermediate Algebra

Question

Solve for \(x\). See Example 3 . $$ \log _{5} x=-2 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{25} \).

Step 1: Understand the Logarithmic Equation

The equation given is \( \log_{5} x = -2 \). This tells us that the logarithm base 5 of \( x \) is equal to \( -2 \). The goal is to find the value of \( x \).

Step 2: Convert Logarithmic Form to Exponential Form

Converting the logarithmic equation to exponential form helps us solve for \( x \). The logarithmic equation \( \log_{5} x = -2 \) can be rewritten as \( x = 5^{-2} \).

Step 3: Calculate the Exponential Value

We now need to calculate \( 5^{-2} \). This means \( x = \frac{1}{5^2} = \frac{1}{25} \).

Key Concepts

Logarithmic Equations

Logarithmic equations involve expressions that incorporate logarithms, which are the inverse operations of exponentials. In simpler words, a logarithm tells us the power to which a number, known as the base, must be raised to get another number. For example, if you see \( \log_{5} x = -2 \), it means that the base 5 must be raised to the power of \(-2\) to result in \(x\). Knowing how to interpret logarithmic expressions is crucial. Logarithmic equations like this are often used to solve problems involving exponential growth or decay, and they can simplify what might be complex multiplications into easier-to-handle addition questions. They are essential in fields such as finance, where they help calculate interest rates, and in science to deal with phenomena like radioactive decay.

Exponential Form

Converting a logarithmic equation to exponential form is an essential step in understanding and solving it. The rule here is straightforward: If you have an equation like \( \log_{b} y = c \), then in exponential form, it becomes \( y = b^c \). This conversion is an eye-opener, as it transforms an unfamiliar logarithmic equation into a more intuitive exponential one. In our example, the equation \( \log_{5} x = -2 \) becomes \( x = 5^{-2} \) upon conversion. This transformation reveals the definition of \(x\) in terms of powers and allows for direct computation. Recognizing this connection helps build a bridge between logarithms and exponential functions, making it easier to handle problems involving growth rates, half-lives, and many forms of scientific calculations. In a practical sense, mastering these conversions means never being stumped by a logarithmic question and always having a tool ready for tackling exponential growth and decay scenarios.

Solving Equations

Solving equations, whether they're linear, quadratic, or involve logarithms, involves finding the values that make the equation true. With logarithmic equations, transforming them into exponential form often simplifies the process. Let's consider the example of \( \log_{5} x = -2 \).Once we convert it to its exponential form, \( x = 5^{-2} \), the solving task becomes calculating \(5^{-2}\). By recognizing \( \text{5}^{-2} \) as the reciprocal of \(5^{2}\), we can directly evaluate it:

• Calculate \( 5^2 \), which is 25.
• Find the reciprocal to get \( x = \frac{1}{25} \).

Following these steps yields the answer: \( x = \frac{1}{25} \). Practicing this approach solidifies the intuition needed for solving various types of equations by converting, calculating, and drawing logical conclusions from exponential transformations. By repeatedly working through these transformations, any student can grow confident in tackling similar algebraic challenges.