Problem 36 - Precalculus with Calculus Previews

Question

Solve the given logarithmic equation. $$ \log _{10} 54-\log _{10} 2=2 \log _{10} x-\log _{10} \sqrt{x} $$

Step-by-Step Solution

Answer

The solution is \( x = 9 \).

Step 1: Apply the properties of logarithm

We have the equation \(\log_{10} 54 - \log_{10} 2 = 2\log_{10} x - \log_{10} \sqrt{x}\). Start by using the properties of logarithms to simplify each side. The properties we will use are: \(\log_{b} a - \log_{b} c = \log_{b} \frac{a}{c}\) and \(k\cdot \log_{b} a = \log_{b} a^k\).

Step 2: Simplify the given equation

Simplify the left-hand side: \(\log_{10} \frac{54}{2} = \log_{10} 27\). Then simplify the right-hand side in two parts: \(2 \log_{10} x = \log_{10} x^2\) and \(\log_{10} \sqrt{x} = \log_{10} x^{1/2}\). Thus, the equation becomes \(\log_{10} 27 = \log_{10} x^2 - \log_{10} x^{1/2}\).

Step 3: Combine the right-hand side

Combine the terms on the right-hand side using the properties of logs: \(\log_{10} x^2 - \log_{10} x^{1/2} = \log_{10} \frac{x^2}{x^{1/2}}\). Note that \(x^2 / x^{1/2} = x^{3/2}\), so the equation is \(\log_{10} 27 = \log_{10} x^{3/2}\).

Step 4: Equate the arguments

Since the logarithms are equal, their arguments must also be equal. Therefore, equate \(27 = x^{3/2}\). Solve for \(x\) by raising both sides to the power of \(\frac{2}{3}\): \(x = 27^{\frac{2}{3}}\).

Step 5: Simplify to find the value of x

Simplifying \(27^{\frac{2}{3}}\), we note that \(27 = 3^3\). Thus, \(27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{3\times\frac{2}{3}} = 3^2 = 9\). Therefore, \(x = 9\).

Key Concepts

Logarithmic Equations

Understanding logarithmic equations is crucial for solving problems that involve unknown exponents. Logarithmic equations often take the form of multiple log terms set equal to each other or to a constant. The key to solving these equations is using the property that if \(\log_b a = \log_b c\), then \(a = c\). This principle allows us to equate the arguments of the logs when their bases are equal.

In the original exercise, the equation involves logarithms with the same base: base 10. Simplifying using logarithmic properties, we can transform and equate terms, making it possible to solve for the unknown variable \(x\). Always double-check that the base of each log is equal, as this step is essential for accurately solving the equation. Once the logs have been simplified and equated, the equation can be solved algebraically.

Properties of Logarithms

Logarithms possess several important properties that are invaluable when simplifying logarithmic expressions or solving logarithmic equations. These properties help transform complex expressions into simpler ones, making it easier to evaluate or solve equations. Here are some key properties utilized in this particular exercise:

• Product Property: \(\log_b (mn) = \log_b m + \log_b n\)
• Quotient Property: \(\log_b \frac{m}{n} = \log_b m - \log_b n\)
• Power Property: \(\log_b (m^n) = n \cdot \log_b m\)

To simplify the initial equation, the quotient rule \(\log_b m - \log_b n = \log_b \frac{m}{n}\) was used on both sides of the equation. Furthermore, the power rule transformed expressions involving exponents. Understanding and applying these properties significantly eases the process and ensures correct simplification of logarithmic equations.

Exponents

Exponents are a fundamental concept in mathematics that pair seamlessly with logarithms. They help in expressing numbers and solving equations. In the context of logarithmic equations, understanding exponents enables proper application of the Power Property of Logarithms.

In this exercise, after simplifying the equation with logarithmic properties, exponents come into play to solve for \(x\). By transforming \(2 \log_{10} x\) and \(\log_{10} \sqrt{x}\) using the Power Property, we end up balancing an equation of exponents. Solving \(x^{3/2} = 27\) involves manipulating the exponents. Simultaneously, understanding exponent rules, such as \((a^m)^n = a^{mn}\), is used to raise \(27\) to the power of \(\frac{2}{3}\) to find \(x\). This exercise emphasized the symmetry between logarithms and exponents, highlighting how mastery of both concepts is essential for solving advanced equations.