Solving Exponential and Logarithm

2 * 4^{3x – 6} – 8 = 120

• x = 3

• x = 4

• x = 2.9

• x = 8

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)

2log₆(x-8)=2

• 14

• 81/8

• 18

• 9969

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

• 8

• 4

• 2

• Cannot be determined

Rewrite 3⁴ = 81 in logarithmic form.

• log₃4 = 81

• log₈₁3 = 4

• log₃81 = 4

• log₄81 = 3

The common logarithm has what base?

• 10

• 0

• e

• -e

Rewrite log_pt = m in exponential form.

• p^t = m

• t^m = p

• m^t = p

• p^m = t

Simplify log₄4^3x

• 3x

• 3

• 4

• 4^3x

e^ln(f)

• f

• e

• e^f

• undefined

log₆(2x + 3) = 3

• x = 106.5

• x = 100

• x = 16

• x = 50