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Chapter 4: Problem 5
Solve each exponential equation by expressing each side as a power of the same base and then equating exponents. $$2^{2 x-1}=32$$
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Without using a calculator, find the exact value of \(\log _{4}\left[\log _{3}\left(\log _{2} 8\right)\right]\).
Use your graphing utility to graph each side of the equation in the same viewing rectangle. Then use the \(x\) -coordinate of the intersection point to find the equation's solution=set. Verify this value by direct substitution into the equation. $$3^{x+1}=9$$
The loudness level of a sound, \(D,\) in decibels, is given by the formula $$D=10 \log \left(10^{12} I\right)$$ where \(I\) is the intensity of the sound, in watts per meter \(^{2} .\) Decibel levels range from \(0,\) a barely audible sound, to \(160,\) a sound resulting in a ruptured eardrum. (Any exposure to sounds of I3 0 decibels or higher puts a person at immediate risk for hearing damage.) What is the decibel level of a normal conversation, \(3.2 \times 10^{-6}\) watt per meter \(^{2} ?\)
Determine whether each statement makes sense or does not make sense, and explain your reasoning. Because the equations \(2^{x}=15\) and \(2^{x}=16\) are similar, I solved them using the same method.
We see from the calculator screen at the bottom of the previous page that a logistic growth model for world population, \(f(x),\) in billions, \(x\) years affer 1949 is $$ f(x)=\frac{12.57}{1+4.11 e^{-0.026 x}} $$ Use this function to solve Exercises \(38-42\) How well does the function model the data showing a world population of 6.1 billion for \(2000 ?\)
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