Problem 83 - College Algebra Essentials

Question

Let \(\log _{b} 2=A\) and \(\log _{b} 3=C .\) Write each expression in terms of \(A\) and \(C\). \(\log _{b} \frac{3}{2}\)

Step-by-Step Solution

Answer

So, \(\log _{b} \frac{3}{2} = C - A\). This is the final result expressed in terms of A and C.

Step 1: Apply quotient property of logarithms

First, apply the quotient property of logarithms to simplify \(\log _{b} \frac{3}{2}\) into separate logarithms. This way the expression becomes: \(\log_b{3} - \log_b{2}\).

Step 2: Substitute the given expressions

Then substitute the given expressions \(\log _{b} 2=A\) and \(\log _{b} 3=C\). Therefore, \(\log_b{3} - \log_b{2}\) will be turned into \(C - A\).

Key Concepts

Quotient Property of Logarithms

Understanding how to simplify logarithmic expressions is fundamental to mastering logarithms. One of the essential properties to learn is the quotient property of logarithms. Let's explore what it really stands for.

The quotient property says that the logarithm of a quotient is equal to the difference between the logarithm of the numerator and the logarithm of the denominator. Mathematically, this can be represented as: \[\begin{equation}\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\end{equation}\]This means that if you have the logarithm of a fraction, you can split it into the subtraction of two separate logarithmic terms, one for the numerator (the top of the fraction) and one for the denominator (the bottom of the fraction).

In our textbook solution, applying this property allowed the expression \(\log_b\frac{3}{2}\) to be expressed as \(\log_b 3 - \log_b 2\), breaking down the compound logarithmic expression into more manageable pieces.

Logarithmic Expressions

Logarithmic expressions can sometimes appear intimidating, but with the right approach, they can be unraveled without much complexity. A logarithmic expression is simply a way to represent an exponent or power. For instance, when we have \(\log_b(x) = y\), it is another way of saying that \(b^y = x\). Here, \(b\) is the base, \(x\) is the argument, and \(y\) is the exponent to which the base must be raised to obtain \(x\).

In the context of our exercise, \(\log_b 2\) and \(\log_b 3\) are given as expressions where the base \(b\) raised to \(A\) or \(C\) would result in 2 and 3, respectively. Breaking down logarithmic expressions into terms of known variables, as we have done by expressing \(\log_b\frac{3}{2}\) using \(A\) and \(C\), helps us simplify and solve equations and can even assist in graphing logarithmic functions by identifying their key characteristics.

Substitution in Logarithms

The concept of substitution is a powerful tool in mathematics, and it's particularly useful when working with logarithms. Substitution involves replacing a variable or an expression with another to simplify the problem or to express the solution in terms of familiar values.

When dealing with logarithmic equations or expressions, as seen in our textbook problem, we may have unspecified variables such as \(A\) and \(C\) that represent certain logarithmic values. If we can express these variables in terms of the logs given, we can then substitute these known values back into the expression to make it more understandable or to facilitate further calculation.

In the step by step solution provided, once we applied the quotient property to the expression \(\log_b\frac{3}{2}\), we used the values given for \(A\) and \(C\) as a substitution to replace \(\log_b 3\) and \(\log_b 2\) respectively, resulting in the much simpler expression \(C - A\). This illustrates the efficiency of substitution in logarithms to transform and solve expressions.