Problem 73 - Algebra 2

Question

Write each logarithm as the quotient of two common logarithms. Do not simplify the quotient. $$ \log _{6}(1-x) $$

Step-by-Step Solution

Answer

\( \log_{6}(1-x) = \frac{\log(1-x)}{\log(6)} \)

Step 1: Understand the logarithm conversion formula

To express the given logarithm as the quotient of two common logarithms, use the change of base formula. The change of base formula states that \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \) for any base \(c\), including the common logarithm base 10.

Step 2: Apply the change of base formula

Apply the change of base formula to the given logarithm \( \log_{6}(1-x) \). Let \(b = 6\) and \(a = 1-x\). Following the formula, we get \( \log_{6}(1-x) = \frac{\log(1-x)}{\log(6)} \) where \(\log\) represents the common logarithm, which is the logarithm with base 10.

Key Concepts

Change of Base Formula

When working with logarithms, you might encounter a situation where you need to convert a logarithm with an unfamiliar base to a more manageable form. The change of base formula is a critical tool for this task. It allows us to rewrite a logarithm in terms of logarithms with a base of our choosing, typically 10 or the natural base 'e', which are more accessible because they are frequently featured on calculators.

Expressing the change of base formula mathematically, we have: \[\begin{equation}\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\end{equation}\]
Here, \(b\) is the original base, \(a\) is the value we're taking the logarithm of, and \(c\) is the new base we're converting to. In practice, the new base \(c\) is often 10, as is the case with common logarithms. The beauty of this formula is that it applies no matter what the base \(c\) is, giving us versatility in mathematical calculations and allowing us to solve logarithmic expressions that might otherwise seem intimidating.

Common Logarithms

Common logarithms are logarithmic functions with a base of 10, denoted simply as \(\log(x)\) without specifying the base. This base is 'common' because of its widespread use in various scientific, engineering, and mathematical contexts. It aligns with our decimal number system, making calculations more intuitive. To explain further, the common logarithm of a number answers the question: 'To what power do we raise 10 to get this number?'.

For instance, \(\log(100)\) queries 'What power of 10 gives us 100?'. The answer is 2, because \(10^2 = 100\). Similarly, if we need to find the common logarithm of a more complex expression, like \(1-x\), and express it without simplifying, we'd write \(\log(1-x)\) directly. By default, the base is understood to be 10, underscoring the 'common' nature of this logarithm.

Logarithmic Expressions

Logarithmic expressions involve the logarithm function, which represent the inverse operations of exponentiation. These expressions can sometimes be complex and involve variables and different bases. Our exercise illustrates how the manipulation of such expressions can be simplified using the change of base formula.

Working with logarithmic expressions requires understanding their properties and operations—like product, quotient, and power rules which allow us to simplify and solve them in much the same way one handles algebraic expressions. For example, \(\log_{6}(1-x)\) may look daunting at first, but with the change of base formula, we're able to unpack it into known components involving common logarithms: \[\log_{6}(1-x) = \frac{\log(1-x)}{\log(6)}\]
This conversion does not alter the value of the expression; it merely presents it in a form that is more convenient to work with—especially when the tools at hand, such as calculators, favour common logarithms.