Find the value of yy in each equation. explain how you determined the value of yy.
a. y = log2(2^2)
b. y = log2(2^5)
c. y = log2(2^−1)
d. y = log2(2^x)

a) 2

b) 5

c) -1

d) x

Step-by-step explanation:

We know the properties of log function as:

1) log(AB) = log(A) + log(B)

2)

3) log(aᵇ) = b × log(a)

also,

4)

Given:

a. y = log₂(2²)

thus, using 3

y = 2log₂(2)

or using 4

y = 2 ×

or

y = 2 × 1 = 2

b. y = log₂(2⁵)

thus, using 3

y =  5 × log₂(2)

or using 4

y = 5 ×

or

y = 5 × 1 = 5

c. y = log₂(2⁻¹)

thus, using 3

y =  -1 × log₂(2)

or using 4

y = -1 ×

or

y = -1 × 1 = -1

d. y = log₂(2ˣ)

thus, using 3

y =  x × log₂(2)

or using 4

y = x ×

or

y = x × 1 = x

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.

a precentage of 6th graders. the population is all 6th graders. let me know if i'm wrong, i hope this ! : )

step-by-step explanation: