Algebra Quiz: Logarithmic Expressions And Equations Test

Math

Grade 11th

CCSS

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| By Albert Melchor

Albert Melchor

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Quizzes Created: 17 | Total Attempts: 12,896

| Attempts: 155 | Questions: 25

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1) Find an equivalent for the given equation.

125 to the power of x equals 5 to the power of x plus 3 end exponent

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About This Quiz

Algebra Quiz: Logarithmic Expressions And Equations Test - Quiz

Ready to tackle the complexities of logarithmic expressions and equations? The Algebra Quiz: Logarithmic Expressions and Equations Test challenges your understanding of key logarithmic concepts, from simplifying expressions to solving equations with precision.

The quiz presents a range of problems that delve into the core properties and rules of... see morelogarithms, testing your ability to navigate both basic and advanced levels. This quiz has questions that will deepen your grasp of algebraic principles and strengthen your problem-solving skills. Take the quiz now to find out how well you can master these logarithmic challenges. Are you ready to put your math knowledge to the test? see less

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2) Which is the best first step to solving the equation?

2 to the power of 3 x plus 1 end exponent equals 54

None of these

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3) Write

4 to the power of 5 equals 1024

in logarithmic form

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4) Find an equivalent for the given equation.

81 to the power of x equals 27 to the power of x plus 3 end exponent

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5) Find an equivalent for the given equation.

3 to the power of 2 x plus 1 end exponent equals 1 over 27

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6) Which is the best first step in solving the equation?

3 to the power of 2 x plus 1 end exponent equals 54

None of these

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7) Solve the equation.

log subscript 3 space left parenthesis 3 x plus 10 right parenthesis space equals space log subscript 3 space x squared

And

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8) Find an equivalent for the given equation.

7 to the power of 4 x minus 5 end exponent equals 1 over 343

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9) Solve the equation.

log subscript 7 space 54 minus log subscript 7 space x space equals space log subscript 7 space 9

X = 4

X = 16

X = 6

X = 9

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The given equation has multiple values of x. However, the correct answer is x = 6 because it is the only value that is listed in the options provided. The other values (4, 16, and 9) are not listed as possible solutions, so they are not the correct answers.

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10) The Allen Electric Company has a piece of machinery valued at $60,000. It depreciates at 20% per year. After how many years will the value have depreciated to $15,000? Use

A equals a left parenthesis 1 plus-or-minus r right parenthesis to the power of t

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The value of the machinery depreciates at a rate of 20% per year. To find out how many years it will take for the value to depreciate to $15,000, we need to calculate the depreciation amount per year. The depreciation amount per year can be found by multiplying the initial value of the machinery ($60,000) by the depreciation rate (20% or 0.2). The depreciation amount per year is $12,000. To find out how many years it will take for the value to depreciate to $15,000, we divide the difference between the initial value and the desired value ($60,000 - $15,000 = $45,000) by the depreciation amount per year ($12,000). Therefore, it will take approximately 3.75 years for the value to depreciate to $15,000.

Explanation

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11) The Allen Electric Company has a piece of machinery valued at $45,000. It depreciates at 20% per year. After how many years will the value have depreciated to $15,000? Use

A equals a left parenthesis 1 plus-or-minus r right parenthesis to the power of t

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The value of the machinery decreases by 20% each year. To find out how many years it takes for the value to depreciate to $15,000, we can set up an equation. Let x represent the number of years. The equation would be: $45,000 * (1 - 0.20)^x = $15,000. By solving this equation, we can find the value of x, which represents the number of years it takes for the machinery to depreciate to $15,000.

Explanation

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12) Jason invests $1200 in an account that pays 3.5% interest compounded continuously. To the nearest cent, how much will be in the account at the end of 2 years? Use

A equals P e to the power of r t end exponent

$1287.01

$2416.50

$2485.49

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The formula for calculating the amount in an account with continuous compounding is A = P * e^(rt), where A is the final amount, P is the principal amount, e is Euler's number (approximately 2.71828), r is the interest rate, and t is the time in years. Plugging in the values, we get A = 1200 * e^(0.035 * 2) = 1200 * e^(0.07) ≈ $1287.01. Therefore, the correct answer is $1287.01.

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13) Which is the best first step in solving the equation?

350 equals 10 to the power of 0.5 x end exponent

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14) Solve the equation.

log subscript 7 space 36 minus log subscript 7 space x equals log subscript 7 space 4

X = 9

X = 32

X = 16

X = 3

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The given equation is x = 9. This means that the value of x is equal to 9.

Explanation

The given equation is x = 9. This means that the value of x is equal to 9.

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15) Express as a single logarithm:

3 space log subscript 5 space x space plus space log subscript 5 space y space minus space 2 space log subscript 5 space w

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16) Using the properties of logs, write in expanded form:

log space open parentheses fraction numerator 2 a b over denominator c cubed end fraction close parentheses

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17) Which is the best first step to solving the equation?

450 equals 10 to the power of 0.8 x end exponent

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18) Solve for x.

log subscript 5 space x squared equals 4

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19) Solve the equation.

log subscript 3 space left parenthesis 3 x plus 4 right parenthesis space equals space log subscript 3 space x squared

and

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20) Solve the equation:

log subscript 2 space left parenthesis 3 x plus 1 right parenthesis space minus space log subscript 2 space left parenthesis x minus 3 right parenthesis space equals space 3

X = 4

X = 3

X = 2

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21) The number of bacteria of a certain type can increase from 80 to 164 in 3 hours. Find the value of k in the general formula for growth and decay, $y equals n e to the power of k t end exponent$

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The general formula for growth and decay is given by the equation: N(t) = N0 * e^(kt), where N(t) is the number of bacteria at time t, N0 is the initial number of bacteria, e is the base of the natural logarithm, and k is the growth or decay constant. In this case, we know that the initial number of bacteria is 80 and the number of bacteria after 3 hours is 164. Plugging these values into the formula, we can solve for k.

Explanation

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22) Jason invests $1000 in an account that pays 2.5% interest compounded continuously. To the nearest cent, how much will be in the account at the end of 2 years? Use

A equals P e to the power of r t end exponent

$1051.27

$2050.63

$1648.72

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Jason invests $1000 in an account that pays 2.5% interest compounded continuously. Continuous compounding means that the interest is constantly being added to the account balance. To calculate the amount in the account at the end of 2 years, we can use the formula A = P * e^(rt), where A is the final amount, P is the principal (initial investment), e is the base of the natural logarithm, r is the interest rate, and t is the time in years. Plugging in the values, we get A = 1000 * e^(0.025 * 2) ≈ $1051.27.

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23) Using the properties of logs, write in expanded form:

ln space open square brackets open parentheses x minus 4 close parentheses open parentheses 2 x plus 5 close parentheses close square brackets squared

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24) The number of bacteria of a certain type can increase from 30 to 254 in 3 hours. Find the value of k in the general formula for growth and decay, $y equals n e to the power of k t end exponent$