Rewriting Logarithms to Exponential Form

log₂32 = 5

• 2^-5 = 32

• 2³² = 5

• 2⁵ = 32

• 32⁵ = 2

Change to Exponential Form:

log₆ 36 = 2

• 2⁶=36

• 6²=36

• 36²=6

• 36⁶=2

Rewrite log_pt = m in exponential form.

• p^t = m

• t^m = p

• m^t = p

• p^m = t

Write log₂ 0.25 = -2 in exponential form

• -2²  = 0.25

• -2^0.25  = 2

• 2^-2  = 0.25

• No correct answer

Rewrite 1.6^(x+2) = 14.1

• log_1.614.1 = x+2

• log_1.6(x+2) = 14.1

• log_14.1(x+2) = 1.6

• log_(x+2)1.6 = 14.1

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

Solve for x:

• 1

• 2

• .5

• 4

log₂(¹/₈) = -3

• 2^-3 = ¹/₈

• 2^1/8 = -3

• -3² = ¹/₈

• -3^1/8 = 2

Change to Exponential Form:
log₆36 = 2

• 2⁶=36

• 6²=36

• 36²=6

• 36⁶=2

log(10,000)=4.

• e⁴=10,000

• 4^10,000=10

• 10⁴=10,000

• 0⁴=10