Chapter 4: Problem 45
Solve the logarithmic equation and eliminate any extraneous solutions. If there are no solutions, so state. $$\log x+\log (x+3)=1$$
Short Answer
Expert verified
The solution to the logarithmic equation \(\log x + \log (x+3) = 1\) is \(x = 2\).
Step by step solution
01
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 \(\log(x(x+3))=1\).
02
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 \(10 = x^2 + 3x\).
03
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 \(x = 2\).
04
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 \(x = 2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).