Answer: 
We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.
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Explanation:
It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.
I'm going to use these three log rules, which apply to any base.
- log(A) + log(B) = log(A*B)
- log(A) - log(B) = log(A/B)
- B*log(A) = log(A^B)
From there, we can then say the following:

The answer is 10/12
i hope this help you
To solve this, add 17 to both sides:
x = -25 + 17
x = -8
Answer:
Well, I don't know how to graph it for you, but the slope is -2/3 and the point is (-9, 5). So, when you graph it, from the point (-9, 5), you can go 3 units to the right and 2 units down.
Answer:
Step-by-step explanation:
this person got it wrong, if your on USATESTPREP its 5,8 your welcome$$$