How to calculate log2 6 (2 is subscript)

How to calculate log2 6 (2 is subscript)

Log2 (subscript) 6 = log2 (subscript) (2x3)
=Log2 (subscript) 2 + log2 (subscript) 3
=1 + log2 (subscript) 3
The subscript is written in word, but it doesn't show here
PS: you should look at the front of the book and the examples

log2(32)-8lg1+log√3(√3)
The results of calculation

Solution
log2(32)-8lg1+log√3(√3)
=log2(2^5)-8×0+1
=5-0+1
=6

Compare the size, 1 / 3 power of log2 with 1 / 3 power of log2 (1 / 2), 1 / 3 power of negative 2 and 1 / 3 power of (3 / 1)

1 / 3 power of log2 > 1 / 3 power of log2
(1 / 2) negative 2 molecule 1 power > (3 molecule 1) 3 molecule 1 power

The image of function f (x) = (1-cosx) SiNx in [- π, π] is approximately
Let me ask you how the image is drawn, let alone differential. It can be calculated that it is an odd function, but why is there a section near the origin of the image on the x-axis very close to the x-axis
It is the ninth choice of mathematics in 2013 national college entrance examination volume 1

First of all, compare the changes of the image of the functions: y = x ^ (1 / 2), y = x, y = x & # 178;, y = x & # 179; near the origin. Do you find that the higher the power exponent is, the closer it is to the X-axis?
F (x) = (1-cosx) SiNx = 4sin & # 179; (x / 2) cos (x / 2) is similar to the above functions

The equation of a straight line passing through point P (2,3) with opposite intercept on two coordinate axes is______ ．

When the straight line passes through the origin, because the slope is 3 − 02 − 0 = 32, the equation of the straight line is y = 32x, that is, 3x-2y = 0. When the straight line does not pass through the origin, let the equation be XA + y − a = 1, substituting the point P (2,3) to get a = - 1, so the equation of the straight line is X-Y + 1 = 0, so the answer is 3x-2y = 0, or X-Y + 1 = 0

How to calculate 15.35 - (5.35 + 7.2) with a simple method?

15.35-(5.35+7.2)
=15.35-5.35-7.2
=10-7.2
=2.8

Given the line y = - 3x + 6 and y = x, find the area of the triangle formed by them and Y axis

Y = - 3x + 6 passing through point (0,6) other intersection and point (1.5,1.5)
So the area is 6 * 1.5 / 2 = 4.5

Given that f (x) = 4 + 1 / x2 under the root sign, the sum of the first n terms of the sequence {an} is Sn, the point PN (an, - 1 / an + 1) (n belongs to n *) is on the curve y = f (x), and A1 = 1, an > 0
Given that f (x) = 4 + 1 / x2 under the root sign, the sum of the first n terms of the sequence {an} is Sn, and the point PN (an, 1 / an + 1) (n belongs to n *) is on the curve y = f (x), and A1 = 1, an > 0
(1) Finding the general term formula of sequence {an}
(2) The first term B1 of the sequence {BN} is 1, the sum of the first n terms is TN, and TN + 1 / A2N = TN / A2N + 1 + 16n2-8n-3. Set the value of B1 so that the sequence {BN} is an arithmetic sequence
(3) To prove 4N + 1-1 under Sn greater than 1 / 2 radical

(1) (4 + 1 / an ^ 2) = 1 / a (n + 1) so 4 + 1 / an ^ 2 = 1 / a (n + 1) ^ 2 so {1 / an ^ 2} the first term 1 / A1 ^ 2 = 1, the arithmetic sequence with common ratio 4 1 / an ^ 2 = 4n-3an = √ [1 / (4n-3)] (2)? The question, B1 = 1, how to set? (3) 1

The power of 2A + 3 of X multiplied by the power of 2B + 1 of Y under the root sign, X and y are greater than 0

Root sign (2a + 3 power of X multiplied by 2B + 1 power of Y)
=√ (2a power of X * X & # 179; * 2B power of Y * y)
=The a power of X * the B power of X * y √ XY
=(a + 1) power of X * B power of Y * √ (XY)

It is known that P: exists that x0 belongs to R, mx02 + 2 ≤ 0, and any x belongs to R, x2-2mx + 1 > 0. If P or q are false propositions, then the value range of real number M

P is true: m (x0) &# 178; + 2 ≤ 0 → m (x0) &# 178; ≤ - 2 → m ≤ 0
Q is true: X & # 178; - 2mx + 1 ＞ 0, → △ = (- 2m) &# 178; - 4 ＜ 0, the solution is - 1 ＜ m ＜ 1
So when q is true and P is true, the value range of M is: - 1 < m ≤ 0
So if P or q is a false proposition, the value range of real number m is m ≤ - 1 or M > 0