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Using the One-to-One Property In Exercises \(73-76,\) use the One-to-One Property to solve the equation for \(x\). $$\ln (x+4)=\ln 12$

Short Answer

Expert verified
The solution to the equation \(\ln (x+4)=\ln 12\) is \(x = 8\).

Step by step solution

01

Identify the One-to-One property

The One-to-One property of logarithms signifies that if \(\ln a = \ln b\), then \(a = b\). Therefore, the equation \(\ln (x+4) = \ln 12\) gives that the arguments on either side of the equation must be equal.
02

Apply the One-to-One property

By applying the One-to-One property, we set the arguments of both sides of the equation equal to each other. We get \(x + 4 = 12\).
03

Solve for x

Subtract 4 from both sides of the equation \(x + 4 = 12\) to get \(x = 12 - 4\). Thus, we get \(x = 8\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Equations
Logarithmic equations are those which include variables inside a logarithm expression. They typically have the general form of \(\log_b(a) = c\), where \(b\) is the base of the logarithm. The key to solving these equations lies in a few important properties of logarithms, such as the One-to-One Property.

This property is fundamental when both sides of the equation have a logarithm with the same base. If two logs are equal, their arguments must be equal as well. For example, if \(\ln(x+4) = \ln(12)\), then the arguments inside the logarithm must also be equal, meaning \(x+4 = 12\).

By understanding and applying these properties, you can solve logarithmic equations efficiently. Remember that familiarity with properties such as the product, quotient, and power rules of logarithms can also help in breaking down more complex logarithmic equations.
Solving Equations
Solving equations is all about finding the value or values of the unknown variable that make the equation true. When faced with an equation, the goal is to isolate the variable on one side of the equation using various algebraic techniques.

For logarithmic equations, we often utilize specific properties, such as the One-to-One Property. Once the logs are simplified and the same on both sides, you can drop them, solving for the variable using basic operations.

In our exercise, once the One-to-One Property revealed \(x + 4 = 12\), it was a simple matter to subtract 4 from both sides to solve for \(x\). This process of isolation is a cornerstone of solving equations in general and highlights the importance of a step-by-step approach.
Precalculus
Precalculus serves as the bridge between algebra and calculus. It introduces concepts and skills that are foundational for calculus, including a deeper understanding of functions, exponential and logarithmic relations, sequences, and series.

The study of logarithms in precalculus is pivotal, as they are extensively used in calculus. Logarithms transform multiplicative processes into additive processes, which simplifies many types of problems, particularly in calculus.

By mastering logarithmic equations and their properties in precalculus, students prepare themselves for more advanced concepts, like derivatives and integrals in calculus, where logs frequently appear. Understanding these precalculus concepts ensures a smoother transition and a stronger mathematical foundation.

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Most popular questions from this chapter

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\log 8 x-\log (1+\sqrt{x})=2$$

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$5^{x}=212$$

Writing Explain why \(\log _{a} x\) is defined only for \(01\) Why vasylehutterstock com

Graphical Analysis Use a graphing utility to graph \(f\) and \(g\) in the same viewing window and determine which is increasing at the greater rate as \(x\) approaches + \(\infty\). What can you conclude about the rate of growth of the natural logarithmic function? (a) \(f(x)=\ln x, \quad g(x)=\sqrt{x}\) (b) \(f(x)=\ln x, \quad g(x)=\sqrt[4]{x}\)

Forensics At 8: 30 A.M., a coroner went to the home of a person who had died during the night. In order to estimate the time of death, the coroner took the person's temperature twice. At 9: 00 A.M. the temperature was \(85.7^{\circ} \mathrm{F},\) and at 11: 00 A.M. the temperature was \(82.8^{\circ} \mathrm{F}\). From these two temperatures, the coroner was able to determine that the time elapsed since death and the body temperature were related by the formula $$t=-10 \ln \frac{T-70}{98.6-70}$$ where \(t\) is the time in hours elapsed since the person died and \(T\) is the temperature (in degrees Fahrenheit) of the person's body. (This formula comes from a general cooling principle called Newton's Law of Cooling. It uses the assumptions that the person had a normal body temperature of \(98.6^{\circ} \mathrm{F}\) at death and that the room temperature was a constant \(70^{\circ} \mathrm{F}\).) Use the formula to estimate the time of death of the person.

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