log2√36 - log2√72

Question

I am not sure how to do this question. I need an explanation.

Answer 1

You will have to know the 3 prime properties of logs

1. log_k (AB) = log_k A + log_k B
2. log_k(A/B) = log_k A - log_k B
3. log_k (A^n) = n log_k A
where k is any positive number , k ≠ 1

so log₂√36 - log₂ log_{2</sub√72
= log2 (√36/√72)
= log2 √(36/72)
= log2 √(1/2)
= log2 2^(-1/2)
= (-1/2) log2 2
= (-1/2)(1)
= -1/2}

Answer 2

log(√x) = 1/2 log(x)
since log(x^n) = n log(x)

So, what you have is
√36 = 6, so

√72 = 6√2

So, assuming by log2 you mean log₂,

log₂6 + log₂6 + log₂2
but log₂2 = 1, so that is
= 2log₂6 + 1/2

Now, what is log₂6?
You don't have a log₂ button on your calculator, but you can easily get it online.

Or, you can change base, remembering that

log₂6 = log₁₀6/log₁₀2

Answer 3

Oops. I added, instead of subtracting. The final answer, of course, is -1/2, as shown above.