I don't understand how log2 √(1/2) turned into log2 2^(-1/2).
Quote:
You will have to know the 3 prime properties of logs
1. logk (AB) = logk A + logk B
2. logk(A/B) = logk A - logk B
3. logk (A^n) = n logk A
where k is any positive number , k ≠ 1
so log2√36 - log2 log2</sub√72
= log2 (√36/√72)
= log2 √(36/72)
= log2 √(1/2)
= log2 2^(-1/2)
= (-1/2) log2 2
= (-1/2)(1)
= -1/2
It would be so much easier if we could identify you by some name other than Anonymous.
To understand how log2 √(1/2) turned into log2 2^(-1/2), let's break it down step by step.
1. First, we have log2 √(1/2).
2. We can simplify this by recognizing that the square root of 1/2 can be written as (1/2) raised to the power of 1/2. So, we now have log2 (1/2)^(1/2).
3. Using the property of logarithms that says logk (A^n) = n logk A, where k is any positive number and k ≠ 1, we can bring down the exponent 1/2 as: (1/2) log2 (1/2).
4. Now, let's simplify further. We know that (1/2) can be written as 2^(-1). So, we have (2^(-1))log2 (1/2).
5. Again, using the property of logarithms logk (AB) = logk A + logk B, we can separate the logarithms: (-1) log2 2 + log2 (1/2).
6. We know that log2 2 equals 1, so now we have (-1) + log2 (1/2).
7. We can simplify further by recognizing that 2^(-1) equals 1/2, so (-1) + log2 (1/2) becomes log2 (1/2).
8. Finally, log2 (1/2) can be rewritten as log2 2^(-1/2).
So, log2 √(1/2) is equal to log2 2^(-1/2).