Problem 64 - Precalculus with Limits

Question

Using Properties of Logarithms In Exercises \(59-66\) , approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562, \log _{b} 3 \approx 0.5646,\) and \(\log _{b} 5 \approx 0.8271\). $$\log _{b}(2 b)^{-2}$$

Step-by-Step Solution

Answer

The approximate value of \(\log_b{(2b)^{-2}}\) is \(-2.7124\)

Step 1: Apply the Power Rule of Logarithms

The power rule states that \(log_b(x^n) = n * log_b(x)\). Therefore, the \(^{-2}\) power in \(\log_b{(2b)^{-2}}\) can be moved in front of the logarithm: \(-2 * \log_b(2b)\).

Step 2: Apply the Product Rule of Logarithms

The product rule says that \(log_b(xy) = log_b(x) + log_b(y)\). Applying this rule, the \(-2 * \log_b(2b)\) can be simplified into \(-2 * (\log_b(2) + \log_b(b))\)

Step 3: Apply the Special Case When Base Equals the Argument

When the base of the logarithm and the argument are the same, the value is 1. Thus, \(log_b(b) = 1\). The expression becomes \(-2 * (\log_b(2) + 1)\)

Step 4: Substitute the Given Value

Using the given \(log_b(2) \approx 0.3562\), substitute this value into the expression. Thus, \(-2 * (0.3562 + 1)\)

Step 5: Calculate the Final Result

After evaluating the expression, by multiplying and adding, the result should be approximately \(-2.7124\)

Key Concepts

Power Rule of Logarithms

Understanding how to manipulate exponents in logarithms can be a game-changer in simplifying complex expressions. The Power Rule is a handy tool in this regard. Let's demystify how it works: consider an expression like \(\log_b(x^n)\), where \((b)\) is the base of the logarithm, \((x)\) is the argument, and \((n)\) is the exponent. The rule allows us to 'bring down' the exponent to the front of the logarithm, thereby transforming our expression to \((n \cdot \log_b(x))\).

When applied to our specific problem, the \(\log_b((2b)^{-2})\) simplifies by moving the \(^{-2}\) and becoming \((-2 \cdot \log_b(2b))\). This step is crucial because it turns a potentially intimidating operation into a much more manageable one, more amenable to further simplification. Remember, this rule consistently works as long as the exponent applies to the entire argument of the logarithm.

Product Rule of Logarithms

The Product Rule shines when we encounter the logarithm of a product. It's like finding a secret passage that splits a single, complex path into two simpler roads. Mathematically, the rule states that for any two positive numbers \((x)\) and \((y)\), the expression \((\log_b(xy)\)) can be rewritten as \((\log_b(x) + \log_b(y))\).

Applying this principle to our example, the expression \((-2 \cdot \log_b(2b))\) is neatly divided into two separate logs: \((-2 \cdot (\log_b(2) + \log_b(b)))\). It's essential to approach this step-by-step; first separate, then simplify. The product rule helps disentangle factors within a logarithmic function, allowing for the individual evaluation of each—subsequently enabling us to plug in known values or make further simplifications as needed.

Logarithm Base Equals Argument

In the curious world of logarithms, there's a unique scenario where the base and the argument of the logarithm are identical. In such cases, the logarithm simplifies to exactly 1. This is no arbitrary coincidence; it's because the logarithm asks us a fundamental question: 'To what exponent must we raise the base to get the argument?' And if the base and argument are the same, the answer is straightforward—just \((1)\), as in \(b^1 = b\).

So, in our earlier step, when we applied this concept to \((\log_b(b))\), it resulted in 1. This simplification is a perfect example of how an expression that looks complicated can hide a simple truth. It ultimately allowed us to replace \((\log_b(b))\) with 1 in our problem, streamlining our calculation significantly. Recognizing this special case, where the logarithm base equals its argument, can lead to immediate simplification and is a powerful tool in a mathematician's toolkit.