Exam 4: Exponential and Logarithmic Functions

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Solve the equation. Write the solution set with the exact values given in terms of natural or common logarithms. Also give approximate solutions to 4 decimal places, if necessary. - 33x3=25x43 ^ { 3 x - 3 } = 2 ^ { 5 x - 4 }

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Use the definition of a one-to-one function to determine if the function is one-to-one. - f(x)=x4f ( x ) = | x - 4 |

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Approximate f(x)=lnxf ( x ) = \ln x for the given value of x. Round to four decimal places. -f (460)

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Write the logarithmic expression as a single logarithm with coefficient 1, and simplify as much as possible. - 5logby+3logbz5 \log _ { b } y + 3 \log _ { b } z

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Choose the one alternative that best completes the statement or answers the question. Solve for the indicated variable. - N=N0e0.0361tN = N _ { 0 } e ^ { - 0.0361 t } for tt lnNN0\ln \frac { N } { N _ { 0 } }

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Solve the equation. Write the solution set with the exact values given in terms of natural or common logarithms. Also give approximate solutions to 4 decimal places, if necessary. - 4e5m52=144 e ^ { 5 m - 5 } - 2 = 14

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Write the word or phrase that best completes each statement or answers the question. Provide the missing information. -The change-of-base formula indicates that logb x can be written as a ratio of logarithms with base a as: logbx=\log _ { b } x = \frac { \square } { \square }

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Determine if the statement is true or false. - ln10=1loge\ln 10 = \frac { 1 } { \log e }

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Simplify the expression. - log0.00001\log 0.00001

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Solve the problem. - $14,000\$ 14,000 is invested at 6%6 \% interest compounded quarterly and grows to $16,738.65\$ 16,738.65 . For how long was the money invested? Round to the nearest year.

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Solve the problem. -The population of a country on January 1, 2000, is 20.720.7 million and on January 1, 2010, it has risen to 22.622.6 million. Write a function of the form P(t)=P0ertP ( t ) = P _ { 0 } e ^ { r t } to model the population P(t)P ( t ) (in millions) tt years after January 1, 2000. Then use the model to predict the population of the country on January 1,2,0171,2,017 . round to the nearest hundred thousand.

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Solve the equation. - log2(72x)log2(x+96)=3\log _ { 2 } ( 72 x ) - \log _ { 2 } ( x + 96 ) = 3

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Solve the equation. - e2x14ex+9=0e ^ { 2 x } - 14 e ^ { x } + 9 = 0

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Write the logarithmic expression as a single logarithm with coefficient 1, and simplify as much as possible. - log4112log47\log _ { 4 } 112 - \log _ { 4 } 7

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Solve the problem. -Given that log50.6990\log 5 \approx 0.6990 and log90.9542\log 9 \approx 0.9542 , use the properties of logarithms to approximate the following. Do not use a calculator. log581\log \frac { 5 } { 81 }

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Solve the problem. -If $18,000\$ 18,000 is put aside in a money market account with interest reinvested monthly at 2.3%2.3 \% , find the time required for the account to earn $2,000\$ 2,000 . Round to the nearest month. Use the model A=PA = P (1+rn)nt\left( 1 + \frac { r } { n } \right) ^ { n t } where AA represents the future value of PP dollars invested at an interest rate rr compounded nn times per year for tt years.

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Solve the problem. -After a new product is launched the cumulative sales S(t)S ( t ) (in $1000)t\$ 1000 ) t weeks after launch is given by: S(t)=641+8e0.47tS ( t ) = \frac { 64 } { 1 + 8 e ^ { - 0.47 t } } What is the limiting value in sales?

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Graph the function and write the domain and range in interval notation. - f(x)=(25)xf ( x ) = \left( \frac { 2 } { 5 } \right) ^ { x }

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Solve the problem. -The function P(t)=8(1.0134)tP ( t ) = 8 ( 1.0134 ) ^ { t } represents the population (in millions) of a country tt years after January 1, 2000. Write an equivalent function using base ee ; that is, write a function of the P(t)=P0ertP ( t ) = P _ { 0 } e ^ { r t } . Round rr to 5 decimal places. Also, determine the population for the year 2000 .

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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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