Exponential and Logarithmic Equations - Pre Assessment

To solve, 16^5x = 64^x+7, what would be the correct set-up?

There is more than one correct answer.

• 4^4(5x)=4^16(x+7)

• 8^2(5x)=8^8(x+7)

• 4^2(5x)=4^3(x+7)

• 2^4(5x)=2^6(x+7)

Answer explanation

To solve 16^(5x) = 64^(x+7), we express both sides with the same base. 16 = 4^2 and 64 = 4^3, leading to 4^(2(5x)) = 4^(3(x+7)). Alternatively, using base 2 gives 2^(4(5x)) = 2^(6(x+7)). Both setups are correct.

Tags

CCSS.HSA.CED.A.1

Solve.

9(e^x) - 31 =23

• x = 1.79

• x = 4.21

• x = -3.5

• x = 1.27

Answer explanation

To solve 9(e^x) - 31 = 23, first add 31 to both sides: 9(e^x) = 54. Then divide by 9: e^x = 6. Taking the natural logarithm gives x = ln(6) ≈ 1.79, which matches the correct answer.

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

CCSS.HSF.LE.A.4

Solve.

2(16^x+6) - 8 = 35

• -4.67

• -4.89

• -2.93

• -2.95

Answer explanation

To solve 2(16^(x+6)) - 8 = 35, first add 8 to both sides: 2(16^(x+6)) = 43. Then divide by 2: 16^(x+6) = 21.5. Taking log base 16 gives x+6 = log16(21.5). Solving yields x = -4.89, the correct answer.

Tags

CCSS.HSF.BF.B.5

CCSS.HSF.LE.A.4

Solve:

log₉(x)+log₉(x+2)=log₉(35)

• 5

• -7

• 5, -7

• -7, -13

Answer explanation

Combine the logs: log_9(x(x+2)) = log_9(35). This gives x(x+2) = 35. Solving the quadratic x^2 + 2x - 35 = 0, we find x = 5 or x = -7. Only x = 5 is valid since logs of negative numbers are undefined.

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

CCSS.HSF.BF.B.5

Solve:

log (x - 1) - log (x - 2) = log 5

• 7

• 4/9

• 9/4

• 1/7

Answer explanation

Using the properties of logarithms, we can combine the left side: log((x - 1)/(x - 2)) = log 5. This implies (x - 1)/(x - 2) = 5. Solving gives x - 1 = 5(x - 2), leading to x = 9/4, which is the correct answer.

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

CCSS.HSF.BF.B.5

To solve 8 = 2^5x+7, you would need to re-write 8 as what base?

• 8

• 4

• 2

• Cannot be determined

Answer explanation

To solve the equation, we need to express 8 as a power of 2. Since 8 = 2^3, we rewrite 8 using base 2. Thus, the correct base to rewrite 8 is 2.

Rewrite the exponential as a log.

2^3x+1 = 12

• ln(2)= 3x+1

• log₂(12)= 3x+1

• ln(6)=3x+1

• log₁₂(2)= 3x+1

Answer explanation

To rewrite the exponential equation 2^(3x+1) = 12 as a logarithm, we use the definition of logarithms: log_b(a) = c means b^c = a. Here, log_2(12) = 3x + 1 is the correct transformation.

Tags

CCSS.HSF.BF.B.5

Solve:

log₂(x + 3)=5

• x=2

• x=29

• x=7

• x=32

Answer explanation

To solve log₂(x + 3) = 5, rewrite it in exponential form: x + 3 = 2^5. This simplifies to x + 3 = 32, leading to x = 32 - 3 = 29. Thus, the correct answer is x = 29.

Tags

CCSS.HSF.BF.B.5

Solve:

2log₆(x-8)=2

• 14

• 81/8

• 18

• 9969

Answer explanation

To solve 2log_6(x-8)=2, divide both sides by 2 to get log_6(x-8)=1. This means x-8=6^1, so x-8=6. Adding 8 gives x=14, which is the correct answer.

Tags

CCSS.HSF.BF.B.5

Which is the correct way to rewrite,log₆(x-3)= 2, in order to solve.

• 6² = x-3

• 2⁶ = x - 3

• ln(2) = x-3

• e⁶ = x-3

Answer explanation

To solve log₆(x-3) = 2, we rewrite it in exponential form. This gives us 6² = x - 3, which is the correct choice. The other options do not correctly represent the logarithmic equation.

Tags

CCSS.HSF.BF.B.5