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
the properties of logs are the same as those of exponents.
√2 = 2^1/2
because
√2 * √2 = 2
and 2^a * 2^b = 2^(a+b)
2^1/2 * 2^1/2 = 2^1 = 2
And, 2^-n = 1/2^n
so,
1/√2 = 2^-1/2
To understand how log2 √(1/2) turned into log2 2^(-1/2), let's break it down step by step:
1. We start with log2 √(1/2).
2. The square root (√) can be represented as raising the number to the power of 1/2. So, we can rewrite it as log2 (1/2)^(1/2).
3. We know the property of logarithm that logk (A^n) = n logk A. Using this property, we can rewrite it as (1/2)log2 (1/2).
4. Now, we need to simplify (1/2)log2 (1/2). We know that 2^(-1/2) is equal to 1/2, so we can rewrite it as (1/2)log2 2^(-1/2).
5. According to the property logk (A^n) = n logk A, (1/2)log2 2^(-1/2) can be further simplified as (-1/2)log2 2.
6. Now, we know that log2 2 is equal to 1, so we have (-1/2)(1).
7. Finally, simplifying (-1/2)(1) gives us a result of -1/2.
Therefore, log2 √(1/2) simplifies to -1/2.