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

4+4+2x

or

2x+8

step-by-step explanation:

answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %