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

think the answer would beb

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)