Q:

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)

Accepted Solution

A:
Answer:a) 2b) 5c) -1d) xStep-by-step explanation:We know the properties of log function as:1) log(AB) = log(A) + log(B)2) [tex]\log(\frac{A}{B}) = \log(A)+\log(B)[/tex]3) log(aᵇ) = b × log(a)also,4) [tex]\log_b(a)=\frac{\log(a)}{\log(b)}[/tex]Given:a. y = log₂(2²)thus, using 3y = 2log₂(2)or using 4y = 2 × [tex]\frac{\log(2)}{\log(2)}[/tex]ory = 2 × 1 = 2b. y = log₂(2⁵)thus, using 3y =  5 × log₂(2)or using 4y = 5 × [tex]\frac{\log(2)}{\log(2)}[/tex]ory = 5 × 1 = 5c. y = log₂(2⁻¹)thus, using 3y =  -1 × log₂(2)or using 4y = -1 × [tex]\frac{\log(2)}{\log(2)}[/tex]ory = -1 × 1 = -1d. y = log₂(2ˣ)thus, using 3y =  x × log₂(2)or using 4y = x × [tex]\frac{\log(2)}{\log(2)}[/tex]ory = x × 1 = x