Logarithmic and Exponential Equations

3^{2x – 6}  = 81

• x = log 4
• x = 5
• x = 4
• x = -1

log₂(2) + log₂(8x) = 6

• x = 3
• x = 2
• x = 6
• x = 4

log₆(2x + 3) = 3

• x = 106.5
• x = 100
• x = 16
• x = 50

Tags

CCSS.HSF.BF.B.5

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

4^2(5x)=4^3(x+7) is one way to solve, which is another alternate set-up?

• 4^4(5x)=4^16(x+7)

• 8^2(5x)=8^8(x+7)

• 4^10x=4^3x+20

• 2^4(5x)=2^6(x+7)

Solve:

log(x) + log(x+3) = 1

• -5, 2

• 5, -2

• -5

• 2

Solve:

log₉(x)+log₉(x+2)=log₉(35)

• 5

• -7

• 5, -7

• -7, -13

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

• 8

• 4

• 2

• Cannot be determined

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

Solve:

log₂(x + 3)=5

• x=2

• x=29

• x=7

• x=32

Tags

CCSS.HSF.BF.B.5

Solve:

2log₆(x-8)=2

• 14

• 81/8

• 18

• 9969

Tags

CCSS.HSF.BF.B.5