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Chapter 3: Problem 91

Is it possible for a logarithmic equation to have more than one extraneous solution? Explain.

Short Answer

Expert verified
Yes, it is possible for a logarithmic equation to have more than one extraneous solution, especially if there are multiple parts in the solved equation that could potentially result in the logarithm of a non-positive number.

Step by step solution

01

Definition of Extraneous Solutions

An extraneous solution is a solution that is derived from the method used in solving the equation, but is not a valid solution to the original equation. They typically appear when both sides of an equation are squared during the solution process, introducing possible solutions that were not part of the original set.
02

Understanding Logarithmic Equations

A logarithmic equation is an equation that involves the logarithm of a variable containing expression. Extraneous solutions in logarithmic equations usually arise when the solution includes a logarithm of a non-positive number, as logarithms are only defined for positive numbers.
03

Conclusion

Therefore, yes, a logarithmic equation can have more than one extraneous solution. This may happen if more than one part of the solved equation results in taking the logarithm of a non-positive number.

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Most popular questions from this chapter

Let \(f(x)=\log _{a} x\) and \(g(x)=a^{x},\) where \(a>1\) (a) Let \(a=1.2\) and use a graphing utility to graph the two functions in the same viewing window. What do you observe? Approximate any points of intersection of the two graphs. (b) Determine the value(s) of \(a\) for which the two graphs have one point of intersection. (c) Determine the value(s) of \(a\) for which the two graphs have two points of intersection.

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.

The graph of \(f(x)=\log _{3} x\) contains the point \((27,3)\)

Data Analysis The table shows the time \(t\) (in seconds) required for a car to attain a speed of \(s\) miles per hour from a standing start. $$\begin{array}{|c|c|} \hline \text { Speed, S } & \text { Time, t } \\ \hline 30 & 3.4 \\ 40 & 5.0 \\ 50 & 7.0 \\ 60 & 9.3 \\ 70 & 12.0 \\ 80 & 15.8 \\ 90 & 20.0 \\ \hline \end{array}$$ Two models for these data are as follows. \(t_{1}=40.757+0.556 s-15.817 \ln s\) \(t_{2}=1.2259+0.0023 s^{2}\) (a) Use the regression feature of a graphing utility to find a linear model \(t_{3}\) and an exponential model \(t_{4}\) for the data. (b) Use the graphing utility to graph the data and each model in the same viewing window. (c) Create a table comparing the data with estimates obtained from each model. (d) Use the results of part (c) to find the sum of the absolute values of the differences between the data and the estimated values given by each model. Based on the four sums, which model do you think best fits the data? Explain.

Using the One-to-One Property In Exercises \(73-76,\) use the One-to-One Property to solve the equation for \(x\). $$\ln (x-7)=\ln 7$$
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