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:
Answer:
1
7 ✔️
15
21
Only 7 is a prime number in this question.
Answer: The answer is ¨B¨
Step-by-step explanation:
Answer:
1
+
sec
2
(
x
)
sin
2
(
x
)
=
sec
2
(
x
)
Start on the left side.
1
+
sec
2
(
x
)
sin
2
(
x
)
Convert to sines and cosines.
Tap for more steps...
1
+
1
cos
2
(
x
)
sin
2
(
x
)
Write
sin
2
(
x
)
as a fraction with denominator
1
.
1
+
1
cos
2
(
x
)
⋅
sin
2
(
x
)
1
Combine.
1
+
1
sin
2
(
x
)
cos
2
(
x
)
⋅
1
Multiply
sin
(
x
)
2
by
1
.
1
+
sin
2
(
x
)
cos
2
(
x
)
⋅
1
Multiply
cos
(
x
)
2
by
1
.
1
+
sin
2
(
x
)
cos
2
(
x
)
Apply Pythagorean identity in reverse.
1
+
1
−
cos
2
(
x
)
cos
2
(
x
)
Simplify.
Tap for more steps...
1
cos
2
(
x
)
Now consider the right side of the equation.
sec
2
(
x
)
Convert to sines and cosines.
Tap for more steps...
1
2
cos
2
(
x
)
One to any power is one.
1
cos
2
(
x
)
Because the two sides have been shown to be equivalent, the equation is an identity.
1
+
sec
2
(
x
)
sin
2
(
x
)
=
sec
2
(
x
)
is an identity
Step-by-step explanation:
Answer:
j = 21 and n = 14
Step-by-step explanation:
we have the equations:
6j + n/3 = 134
j/3 + n = 31
54j + 3n = 1206
j + 3n = 93
53j = 1113
j = 21
(21)/3 + n = 31
7 + n = 31
n = 14
