Exam 4: Exponential and Logarithmic Functions

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Choose the one alternative that best completes the statement or answers the question. Evaluate the function at the given value of x. Round to 4 decimal places if necessary. - f(x)=(14)x;f(5)f ( x ) = \left( \frac { 1 } { 4 } \right) ^ { x } ; \quad f ( - 5 )

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Solve the logarithmic equation. - log9t3=log9t22\log _ { 9 } t ^ { 3 } = \log _ { 9 } t ^ { 2 } - 2

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Choose the one alternative that best completes the statement or answers the question. Write the equation in exponential form. - log416=2\log _ { 4 } 16 = 2

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A function defined by y=f(x) (is/is not) y = f ( x ) \text { (is/is not) } ) (is/is not) a one-to-one function if no horizontal line intersects the graph of f in more than one point.

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Solve the equation. - log5(5p+3)+log5p=log514\log _ { 5 } ( 5 p + 3 ) + \log _ { 5 } p = \log _ { 5 } 14

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Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. - log5xlog5(2x+6)=12log54\log _ { 5 } x - \log _ { 5 } ( 2 x + 6 ) = \frac { 1 } { 2 } \log _ { 5 } 4

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Solve the problem. -The formula L=10log(II0)L = 10 \log \left( \frac { I } { I _ { 0 } } \right) gives the loudness of sound LL (in dB\mathrm { dB } ) based on the intensity of the sound (in W/m2\mathrm { W } / \mathrm { m } ^ { 2 } ). The value I0=1012 W/m2I _ { 0 } = 10 ^ { - 12 } \mathrm {~W} / \mathrm { m } ^ { 2 } is the minimal threshold for hearing for mid-frequency sounds. Hearing impairment is often measured according to the minimal sound level (in dB\mathrm { dB } ) detected by an individual for sounds of various frequencies. If the minimum loudness of sound detected by an individual is 60 dB60 \mathrm {~dB} , determine the corresponding intensity of sound.

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Graph the points and from visual inspection, select the model that would best fit the data. Choose from y=mx+b (linear) y=a (exponential) y=a+bx (logarithmic) y= (logistic) Then use a graphing utility to find a function that fits the data. (Hint: For a logistic model, go to STAT, CALC, Logistic.) - x y 4 4.4 5 5.2 9 10.5 14 25.1 20 71.3 24 143.1

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Simplify the expression. - 4log4(x57)4 \log _ { 4 } \left( x ^ { 5 } - 7 \right)

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Determine if the relation defines y as a one-to-one function of x. -Determine if the relation defines y as a one-to-one function of x. -

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Find an equation for the inverse function. - f(x)=8x3f ( x ) = 8 ^ { x } - 3

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Find an equation for the inverse function. - f(x)=ln(x+3)f ( x ) = \ln ( x + 3 )

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Solve the problem. -Given that the domain of a one-to-one function ff is [8,1)[ - 8 , - 1 ) and the range of ff is (8,)( 8 , \infty ) , state the domain and range of f1f ^ { - 1 } .

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Graph the function. - y=log1/3xy = \log _ { 1 / 3 } x

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Write the word or phrase that best completes each statement or answers the question. Provide the missing information. -  As x, the value of (1+1x)x approaches \text { As } x \rightarrow \infty \text {, the value of } \left( 1 + \frac { 1 } { x } \right) ^ { x } \text { approaches } ــــــــــــــ.

(Short Answer)
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Use transformations of the graph y=exy = e ^ { x } to graph the function. Write the domain and range in interval notation. - f(x)=ex+2f ( x ) = e ^ { x + 2 }

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Write the word or phrase that best completes each statement or answers the question. Provide the missing information. - logbb=\log _ { b } b = ـــــــــــــــــــ because b=bb ^ { \square } = b .

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Write the word or phrase that best completes each statement or answers the question. Provide the missing information. -The domain of an exponential function f(x)=bxf ( x ) = b ^ { x } is .

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Graph the points and from visual inspection, select the model that would best fit the data. Choose from y=mx+b (linear) y=a (exponential) y=a+bx (logarithmic) y= (logistic) Then use a graphing utility to find a function that fits the data. (Hint: For a logistic model, go to STAT, CALC, Logistic.) - x y 10 41.2 20 47.5 30 51.2 40 53.8 50 55.9 60 57.5

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Write the word or phrase that best completes each statement or answers the question. Provide the missing information. -  Given a logistic growth function y=c1+aebt, the limiting value of y is \text { Given a logistic growth function } y = \frac { c } { 1 + a e ^ { - b t } } \text {, the limiting value of } y \text { is } ــــــــــــــ

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