Solving Exponentials with Logs

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)

Solve.

9(e^x) - 31 =23

• x = 1.79

• x = 4.21

• x = -3.5

• x = 1.27

Solve.

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

• -4.67

• -4.89

• -2.93

• -2.95

Solve:

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

• 5

• -2

• -5

• 2

Solve:

log₉(x)+log₉(x+2)=log₉(35)

• 5

• -7

• 5, -7

• -7, -13

Solve:

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

• 7

• 4/9

• 9/4

• 1/7

To solve 8 = 2^5x+7, you would need to 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

Solve:

log₂(x + 3)=5

• x=2

• x=29

• x=7

• x=32

Solve:

2log₆(x-8)=2

• 14

• 81/8

• 18

• 9969