Question

Use the properties of logarithms to simplify the given logarithmic expression.\(\log _{5} \frac{1}{250}\)

Short answer

-2.

Step-by-step solution

Express the fraction as multiplication

First, it's useful to express the fraction within the logarithm as a multiplication. The fraction \(\frac{1}{250}\) can be rewritten as \(1*5^{-2}\). Thus, the given expression can be written as \(\log _{5}(1*5^{-2})\).

Apply the logarithm rule for multiplication

Now, apply the logarithm rule for multiplication, \(\log_a(b * c) = \log_a(b) + \log_a(c)\). Therefore, the given expression can be further simplified to \(\log _{5}1 + \log _{5}5^{-2}\).

Simplify the logarithmic expressions

The logarithm of 1 to any base is 0, so \(\log _{5}1\) is 0. The expression \(\log _{5}5^{-2}\) simplifies to \(-2\), because according to the rule \(\log_a(a^c) = c\), logarithm of a number with the base being the same number is the exponent itself. So, the entire expression simplifies to \(0 + -2) = -2.\)

Key concepts

Logarithmic Expressions

Logarithmic expressions can initially appear intimidating, but they are simply a different way to represent exponential relationships. In its most basic form, a logarithm answers the question: 'To what exponent must we raise the base to obtain a certain number?' For example, the expression \( \log_b(x) \) means 'What power must we raise the base \( b \) to get \( x \)?'

In the exercise \( \log _{5} \frac{1}{250} \) we are effectively asking, 'To what power must we raise 5 to get \( \frac{1}{250} \)?' Understanding this fundamental question is crucial for simplifying logarithmic expressions and applying the various logarithm rules that allow us to solve more complex problems.

Simplifying Logarithms

Simplifying logarithms is a fundamental skill in mathematics. To make a logarithmic expression simpler to understand, we often have to manipulate it using the properties of logarithms. For instance, as seen in the exercise solution, breaking down a fraction within a logarithm into its components can be helpful. The fraction \( \frac{1}{250} \) becomes \( 1 \times 5^{-2} \) when rewritten, allowing us to use the multiplication property. This is a practical demonstration of how starting with a complex-looking expression, and utilizing a series of logical steps, leads us to a much more approachable form, like turning \( \log _{5} \frac{1}{250} \) into a sum of simpler logarithms, which ultimately simplifies the task at hand.

Another tip is to remember that \( \log \) of 1 to any base is 0. Breaking the problem down and applying these fundamental simplifications transform what might seem like an overwhelming expression into something much more manageable, leading to clear, concise solutions.

Logarithm Rules

The properties of logarithms, or 'logarithm rules,' are powerful tools that students can use to work through logarithmic expressions. There are several key rules that are particularly useful:

• Product Rule: \( \log_a(b \times c) = \log_a(b) + \log_a(c) \) - This allows us to turn the log of a product into a sum of logs.
• Quotient Rule: \( \log_a(\frac{b}{c}) = \log_a(b) - \log_a(c) \) - Converts the log of a fraction into a difference of logs.
• Power Rule: \( \log_a(b^c) = c \times \log_a(b) \) - Lets you move the exponent in front of the log.
• Change of Base Formula: \( \log_a(b) = \frac{\log_c(b)}{\log_c(a)} \) - Useful for calculating logs in a different base than the one given.
• Logarithm of 1: \( \log_a(1) = 0 \) - The result is always zero no matter the value of \( a \) as long as \( a > 0 \) and \( a eq 1 \)..
• Logarithm of the Base: \( \log_a(a) = 1 \) and more generally, \( \log_a(a^c) = c \) - The log of a number where the base and the number are the same equals the exponent.

In our original problem, we applied the product and power rules to simplify the expression, demonstrating how these rules function more like a set of instructions that, when followed, simplify the process of managing logarithms.