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.
===========================================================
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:
She should have subtracted 12 instead of adding.
x+12=31
-12=-12
x=19
<span>x^2 - 6x = 14
-6
divided by 2 =
-3
then squared =
9
</span><span>x^2 - 6x +9 = 14 +9
(x -3)^2 = 23
answer is B
Source:
http://www.1728.org/quadr2.htm
</span>
first multiply 14•3 = 42
find 2 numbers that when multiplied it equals to 42 and when added it equals to 23, in this case the numbers are 21 & 2
replace the 23d with the 2 numbers
factor out 3d from the expression
then factor out 2 from the expression
factor out (d+7)