Exam 4: Inverse, Exponential, and Logarithmic Functions

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Determine whether or not the function is one-to-one. - f(x)=2x3−5f ( x ) = 2 x ^ { 3 } - 5

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Graph the function. - f(x)=3−xf(x)=3-x  Graph the function. - f(x)=3-x

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Solve the equation. - log⁡(x−1)15=1\log ( x - 1 )^{ 15} = 1

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Solve the problem. -Find the number of years for $8400 to grow to $12,900 at 6% compounded semiannually. Round to the nearest tenth of a year.

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Solve the equation. Round to the nearest thousandth. - 2e(4x+6)=82 \mathrm { e } ^{( 4 x+ 6 )} = 8

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Solve the following graphically. If necessary, round answers to the nearest thousandth. - ex+3=1x+5\mathrm { e } ^ { \mathrm { x } + 3 } = \frac { 1 } { \mathrm { x } + 5 }

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Write the word or phrase that best completes each statement or answers the question. -The graph of y=ex\mathrm { y } = \mathrm { e } ^ { \mathrm { x } } is shown with the coordinates of a point displayed at the bottom of the screen. Write the logarithmic equation associated with the display.  Write the word or phrase that best completes each statement or answers the question. -The graph of  \mathrm { y } = \mathrm { e } ^ { \mathrm { x } }  is shown with the coordinates of a point displayed at the bottom of the screen. Write the logarithmic equation associated with the display.

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Solve the problem. -The logarithmic function f(x)=−100+86ln⁡x\mathrm { f } ( \mathrm { x } ) = - 100 + 86 \ln \mathrm { x } models the number of visitors (in millions) to a certain country's museums, where xx is the number of years since the initial recording of the number of visitors. Use this function to estimate the number of visitors in year 80 . Round to the nearest tenth of a year.

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Solve the problem. -The decay of 667 mg of an isotope is given by A(t)= 667e-0.015t, where t is time in years since the initial amount of 667 mg was present. Find the amount (to the nearest milligram) left after 24 Years.

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Solve the equation. Round to the nearest thousandth. - 3x=173 ^ { x } = 17

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Decide whether the given functions are inverses. - f=\{(3,4),(6,3),(0,8)\} g=\{(3,6),(8,0),(4,4)\}

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Solve the problem. -The half-life of Cesium 134 m134 \mathrm {~m} is 3.03.0 hours. If the formula P=(12)t/3.0\mathrm { P } = \left( \frac { 1 } { 2 } \right) ^ { \mathrm { t } / 3.0 } gives the percent (as a decimal) remaining after time tt (in hours), sketch PP versus tt .  Solve the problem. -The half-life of Cesium  134 \mathrm {~m}  is  3.0  hours. If the formula  \mathrm { P } = \left( \frac { 1 } { 2 } \right) ^ { \mathrm { t } / 3.0 }  gives the percent (as a decimal) remaining after time  t  (in hours), sketch  P  versus  t .

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Solve the problem. -The loudness of a sound can be quantified in units called decibels, where the number of decibels d is given by the formula d=10log⁡II0d = 10 \log \frac { I } { I _ { 0 } } . What is the decibel rating of a sound having an intensity I=100I0I = 100 \mathrm { I } _ { 0 } ?

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Solve the problem. -Suppose that y=2−log⁡(100−x)0.4\mathrm { y } = \frac { 2 - \log ( 100 - \mathrm { x } ) } { 0.4 } can be used to calculate the number of years y\mathrm { y } for x percent of a population of 303 web-footed sparrows to die. Approximate the percentage (to the nearest whole per cent) of web-footed sparrows that died after 2yr2 \mathrm { yr } .

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Write the word or phrase that best completes each statement or answers the question. -A lake is stocked with 452 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 2825 . The population of fish in the lake after time tt , in months, is given by the function P(t)=28251+5.77e−0.28tP ( t ) = \frac { 2825 } { 1 + 5.77 e ^ { - 0.28 t } } . After how many months will the population be 2600 ?

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Decide whether the given functions are inverses. - f=\{(5,13),(6,2),(2,24),(13,5)\} g=\{(13,5),(2,6),(24,2),(5,13)\}

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Graph the exponential function using transformations where appropriate. - f(x)=4x+5f(x)=4^{x+5}  Graph the exponential function using transformations where appropriate. - f(x)=4^{x+5}

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Solve the equation. - log⁡416=x\log _ { 4 } 16 = x

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Use the properties of logarithms to rewrite the expression. Simplify the result if possible. Assume all variables represent positive real numbers. - log⁡6(4611)\log _ { 6 } \left( \frac { 4 \sqrt { 6 } } { 11 } \right) log⁡64+12log⁡66\log _ { 6 } 4 + \frac { 1 } { 2 } \log _ { 6 } 6

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Solve the problem. -A certain noise has an intensity I of 4.01×10−54.01 \times 10 ^ { - 5 } . Given that decibel level D is related to intensity by D=10log⁡(II0)\mathrm { D } = 10 \log \left( \frac { \mathrm { I } } { \mathrm { I } _ { 0 } } \right) , where I0\mathrm { I } _ { 0 } is 10−1210 ^ { - 12 } , determine the decibel level of the noise. Round your answer to the nearest decibel.

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