Problem 117 - Algebra and Trigonometry Real Mathematics, Real People

Question

(a) complete the table to find an interval containing the solution of the equation, (b) use a graphing utility to graph both sides of the equation to estimate the solution, and (c) solve the equation algebraically. Round your results to three decimal places. $$6 \log _{3}(0.5 x)=11$$ $$\begin{array}{|l|l|l|l|l|l|}\hline x & 12 & 13 & 14 & 15 & 16 \\\\\hline 6 \log _{3}(0.5 x) & & & & & \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

The solution of the equation \(6 \log _{3}(0.5 x)=11\) is approximately \(x=16.899\)

Step 1: Completing the Table

Plug each 'x' value into the equation to get the corresponding 'y' values: \(6 \log _{3}(0.5 \times 12)\),\(6 \log _{3}(0.5 \times 13)\),\(6 \log _{3}(0.5 \times 14)\),\(6 \log _{3}(0.5 \times 15)\),\(6 \log _{3}(0.5 \times 16)\). Make note of the value where the output is close to 11, which means the true answer is within that interval.

Step 2: Drawing the Graph

Plot the logarithmic function \(y=6 \log _{3}(0.5x)\) and the equation line \(y=11\) on the same coordinate axis. Locate the point where the two graphs intersect. The 'x' coordinate of that intersection point is the estimated solution.

Step 3: Solving Algebraically

To solve this equation algebraically, first, divide the equation by 6 to get \(\log _{3}(0.5 x)=\frac{11}{6}\) . Then transform the logarithmic equation to exponential form to get \(0.5x=3^{\frac{11}{6}}\). Divide both sides by 0.5 to find the solution for 'x'.

Key Concepts

Logarithmic Functions

A logarithmic function is the inverse of an exponential function. It's written in the form of \( y = \log_b(x) \) where \( b \) is the base, \( x \) is the argument, and \( y \) is the logarithm of \( x \) to the base \( b \). The equation indicates that \( b^y = x \).

In the given exercise, the function \( 6 \log _{3}(0.5 x) \) involves a base of \( 3 \) and an argument of \( 0.5x \) multiplied by \( 6 \). Logarithmic functions only take positive numbers as arguments and their bases must be positive and not equal to \( 1 \). As students work through log equations, it's crucial to remember these properties to avoid potential mistakes. Understanding how to evaluate and manipulate these functions is vital in many fields, including science, engineering, and finance.

Graphing Logarithms

Graphing a logarithmic function can help with understanding its behavior and identifying solutions to logarithmic equations. The graph of a basic logarithmic function \( y = \log_b(x) \) has a vertical asymptote at \( x=0 \) and passes through the point \( (1,0) \) because \( b^0 = 1 \).

When graphing, it is useful to remember that as \( x \) approaches \( 0 \), the \( y \) value will decrease without bound, reflecting the vertical asymptote. For the given problem, graphing the function \( y=6 \log _{3}(0.5x) \) requires a transformation of this basic log function, which involves stretching vertically by a factor of \( 6 \) and horizontally by a factor of \( 0.5 \). Additionally, plotting the constant function \( y=11 \) and finding the intersection with the logarithmic function's graph provides a visual estimate of the solution for \( x \).

Algebraic Methods

To solve logarithmic equations algebraically, one often employs properties of logarithms such as the Product, Quotient, and Power Rules to simplify the expression. However, in certain instances like the one in the exercise, the equation must first be simplified by isolating the logarithmic part, which is accomplished by dividing both sides by \( 6 \) to obtain \( \log _{3}(0.5 x)=\frac{11}{6} \).

After this initial step, applying the definition of the logarithm transforms the equation into its exponential counterpart. Solving the resulting exponential equation typically requires straightforward algebraic manipulation. It's essential to keep in mind that while solving, one must check for extraneous solutions that can arise when applying logarithmic properties.

Exponential Form Transformation

Transforming a logarithmic equation into exponential form is an important technique for finding solutions. This is based on the fact that if \( \log_b(x) = y \), then \( b^y = x \). This fundamental property enables us to switch between the logarithmic and exponential forms seamlessly.

In the exercise, \( \log _{3}(0.5 x)=\frac{11}{6} \) is converted to \( 0.5x=3^{\frac{11}{6}} \) by raising \( 3 \) to the \( \frac{11}{6} \) on both sides of the equation, thus effectively 'canceling' the log and leaving the \( x \) variable to be solved for. The final step involves solving for \( x \) by dividing by \( 0.5 \) which provides the solution for the original problem. Mastery of this transformation is crucial because it demystifies the process of finding \( x \) in log equations.