Problem 52 - Precalculus : Building Concepts and Connections

Question

In Exercises \(47-52,\) let \(b=\log\) k. Write each expression in terms of b. Assume \(k>0\). $$\log \frac{1}{k^{3}}$$

Step-by-Step Solution

Answer

Thus, the expression \(\log \frac{1}{k^{3}}\) can be expressed in terms of \(b\) as \(-3b\).

Step 1: Applying the Rule of Logarithm for Fraction

We begin by turning the fraction inside the logarithm into a difference of two distinct logarithms since \(\log \frac{1}{a}\) can be written as \(- \log a\). So, we have: \(\log \frac{1}{k^{3}} = - \log k^{3}\).

Step 2: Applying the Rule of Logarithm for Exponent

Next we apply the rule \(\log a^{n}=n \log a\). This means that when a logarithm acts upon an exponent, as \(k^{3}\) in our case, the exponent comes out as a multiplier. This gives: \(- \log k^{3} = -3 \log k\).

Step 3: Substituting for b

Since we know that \(b=\log k\), we can substitute this into the expression got in Step 2. So we obtain: \(-3 \log k = -3b\).

Key Concepts

Logarithm Fraction Rule

Understanding the logarithm fraction rule is essential for simplifying complex logarithmic expressions. The rule states that the logarithm of a fraction can be expressed as the difference between the logarithm of the numerator and the logarithm of the denominator. In simpler terms, if you have a fraction inside a logarithm, like \(\log\left(\frac{a}{b}\right)\), you can split it into \(\log(a) - \log(b)\).For the given exercise, the fraction within the logarithm is \(\frac{1}{k^3}\), and according to the fraction rule, it translates to \(\log(1) - \log(k^3)\). Since \(\log(1)\) always equals zero because any number to the power of zero is one, the expression simplifies to \(0 - \log(k^3)\), or just \( -\log(k^3)\).Using this rule effectively requires practice. It simplifies expressions and provides a clear path to further manipulation using other logarithmic identities, making complex calculations more manageable.

Logarithm Exponent Rule

The exponent rule for logarithms is a powerful tool when dealing with exponents within logarithmic functions. It states that the logarithm of a number raised to an exponent is equal to the exponent times the logarithm of the base number. That is, \(\log(a^n) = n\log(a)\).In our exercise, we apply this rule to the term \(\log(k^3)\) to pull the exponent outside the logarithm function. As a result, we get \(3\log(k)\). Then, by applying a negative sign as per the previous operation due to the fraction rule, the expression becomes \( -3\log(k)\).The exponent rule not only makes the computations straightforward but also can be helpful in situations when you have logarithms with variable exponents. It's an efficient method to break down and simplify complex logarithmic expressions involving powers.

Expressing Logarithms in Terms of Variables

Logarithms can often be expressed in terms of variables to simplify expressions and make them more understandable. In the example exercise, we are asked to express the logarithmic function in terms of \(b\), where \(b = \log k\).By applying previous rules, we have transformed the original expression to \( -3\log(k) \). To express this in terms of \(b\), we substitute \(\log(k)\) with \(b\), as defined in the exercise, and the final expression is now \(-3b\).This technique is not only a mere substitution but also a way to represent an expression in various forms depending on the context and simplify calculations. It is a common strategy in precalculus to deal with complex logarithmic equations and can aid in graphing logarithmic functions, understanding their behavior, or even in solving logarithmic equations. Communication of mathematical ideas also becomes more fluid when using variables like \(b\), embodying complex operations.