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:

To get rid of the 2/5 next to the g, multiply both sides by the RECIPROCAL of 2/5, which would be 5/2.
The answer is g=15/10, which simplifies to g=1 5/10 and then further simplifies (wow so much simplifying) to get g=1 1/2.
Answer:
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Step-by-step explanation:
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X+y+z=32
x=z-4
y=x+1
z=x+4
x+x+1+x+4=32
3x+5=32
3x=27
x=9
y=10
z=13
Hope this helps :)