Exponential and Log Solving

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

Tags

CCSS.HSF.BF.B.5

Determine the correct set-up for: log₄(x-1) - log₄(x+7) = log₄6

• log₄(x-1)/(x+7)=log₄(6)

• log₄(x-1)(x+7)=log₄(6)

• log₄(x+7)/(x-1)=log₄(6)

• (x-1)(x+7)=6

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

• 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)

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

• -4.67

• -4.89

• -2.93

• -2.95

Solve:

log₉(x)+log₉(x+2)=log₉(35)

• 5

• -7

• 5, -7

• -7, -13

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

• 8

• 4

• 2

• Cannot be determined

log₂(x + 3)=5

• x=2

• x=29

• x=7

• x=32

Tags

CCSS.HSF.BF.B.5

2log₆(x-8)=2

• 14

• 81/8

• 18

• 9969

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

Tags

CCSS.HSF.BF.B.5

log₈(6-5x) = log₈(3- 4x)

• x = 3

• No Solution

• x = 9

• x = 1