log2√36 - log2√72

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

3 answers

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