Problem 45 - Precalculus : Building Concepts and Connections

Question

Solve the logarithmic equation and eliminate any extraneous solutions. If there are no solutions, so state. $$\log x+\log (x+3)=1$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log x + \log (x+3) = 1\) is \(x = 2\).

Step 1: Apply the Properties of Logarithms

First, use the logarithmic property \(\log a + \log b = \log ab\) to combine the two logarithms on the left side of the equation. This will give \(\log(x(x+3))=1\).

Step 2: Rewrite the Logarithmic Equation as an Exponential Equation

The equation \(\log_b a = c\) is equivalent to the exponential equation \(b^c = a\). With this property, rewrite the equation \(\log(x(x+3))=1\) as \(10^1 = x(x+3)\), which simplifies to \(10 = x^2 + 3x\).

Step 3: Make a Quadratic Equation and Solve

Rearrange the equation above to form a quadratic equation \(x^2 + 3x - 10=0\). Solve this equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). In this case, \(a=1, b=3, c=-10\). Plugging those values in, we obtain \(x = \frac{-3 \pm \sqrt{3^2-4*1*(-10)}}{2*1}\), which simplifies to \(x = \frac{-3 \pm \sqrt{49}}{2}\). From the equation, we obtain \(x = -7\) or \(x = 2\).

Step 4: Check for Extraneous Solutions and State the Final Answer

Check these solutions in the original equation. Substitute \(x = -7\) into the equation and we obtain \(\log(-7) + \log(-7+3)\), which is undefined, making \(x = -7\) an extraneous solution. Substitute \(x = 2\) into the equation and we obtain \(\log(2) + \log(2+3)\), which simplifies to \(1\), making \(x = 2\) a valid solution. Hence, the only solution is \(x = 2\).

Key Concepts

Properties of Logarithms

Logarithmic equations often require a deep understanding of the properties of logarithms to solve effectively. One of the essential properties is the sum-to-product property, which states that the sum of two logarithms with the same base can be combined into a single logarithm:

• Formula: \(\log_a b + \log_a c = \log_a(bc)\)This property allows you to merge separate logarithmic terms into one, making the equation easier to manage.
• Understanding the product property can immensely simplify logarithmic problems. By combining terms, we reduce the complexity from dealing with multiple logs to handling a single log.

In the original problem, we used this property to combine \(\log x + \log(x+3)\) into \(\log(x(x+3))\). This had the advantage of simplifying the left-hand side of the equation before moving on to further solve it.Using properties of logarithms thus serves as a foundational step in solving many logarithmic equations. Remembering these properties and practicing their application can make solving these types of equations much more straightforward.

Exponential Equations

After applying the properties of logarithms, it is often necessary to convert a logarithmic equation into an exponential form to make it more solvable. The conversion between logarithmic and exponential forms is crucial since it sometimes leads to simpler arithmetic solutions.

• The basic conversion rule is that the equation \(\log_b a = c\) can be rewritten as \(b^c = a\).
• This conversion leverages the inherently inverse relationship between logarithms and exponents, allowing you to switch the form of the equation.

In our exercise, we converted \(\log(x(x+3)) = 1\) into an exponential form \(10^1 = x(x+3)\). This conversion was instrumental in translating the logarithmic equation into something familiar—a polynomial equation that can be more easily handled and solved using algebraic techniques. Understanding exponential equations and how to manipulate them is a key skill in resolving equations that originate from logarithmic expressions.

Quadratic Equations

Once a logarithmic equation is transformed into an exponential one, it can sometimes lead to a quadratic equation. Solving quadratic equations is a fundamental skill in algebra, making their understanding crucial when handling transformed equations.

• Quadratic equations are usually in the form \(ax^2 + bx + c = 0\).
• They can often be solved using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

In our case, converting the logarithmic equation ultimately yielded \(x^2 + 3x - 10=0\). By applying the quadratic formula, the possible solutions were found: \(x = -7\) and \(x = 2\). The important step that follows finding the solutions is to check them in the context of the original equation. Some solutions, like \(x = -7\), might not be valid due to the domain restrictions of logarithms. This particular checking resulted in discarding \(x = -7\) due to logarithms being undefined for negative numbers. Recognizing valid solutions from extraneous ones reinforces the depth of understanding required to solve quadratic equations accurately within these contexts.