How to calculate - (6 / 8) * log2 (6 / 8)? How to press log with the computer? I forgot. I can't remember

How to calculate - (6 / 8) * log2 (6 / 8)? How to press log with the computer? I forgot. I can't remember

log2（6/8）=log2（3）+log2（4）=log2（2）+2

F (x) = 1 + F (1 / x) * log2 ^ x, then what is f (2) equal to?

If f (2) = 1 + F (1 / 2) * log2 ^ 2 = 1 + F (1 / 2) and f (1 / 2) = 1 + F (2) * log2 ^ (1 / 2) = 1-f (2), then f (2) = 1, f (1 / 2) = 0

Calculation: log2 (8 × 16) = () + () = ()

Log2 (8) plus log2 (16) equals 3 plus 4 equals 7

Log2 (16 ^ 2 * 64) calculation
I have just learned this question myself. What I want to ask is how to calculate LG8 / LG4
How did log2 in [log2 (8) / log2 (10)] come from,

log2(16^2)+log2(64)=2*4+6=14.
lg8/lg4=[log2(8)/log2(10)]*[log2(10)/log2(4)]=log2(2^3)/log2(2^2)=3/2.

To prove that the sum of squares of a and B is greater than or equal to 2Ab, we can use the complete square of A-B to prove it, but how to use the square of (a + b) to prove it

(a + b) (a + b) = the square of a plus the square of B plus 2 ab. if the square of a + B is greater than 0, then a square plus b square plus 2 ab is greater than 0
a. The sum of squares of B is greater than or equal to 2A

F (x) has a second order continuous derivative, f (0) = 0. It is proved that G (x) has a derivative from negative infinity to positive infinity
As the title

When x is not equal to zero, G (x) = f (x) / x, obviously f (x) has second order continuous derivative, 1 / X is also differentiable, so G '(x) = [XF' (x) - f (x)] / x ^ 2. When x is not equal to 0, f '(x) is also continuous because f (x) has second order continuous derivative, obviously 1 / x ^ 2 is also continuous

It must be solved by proportion! An airplane travels 720 kilometers per hour, flies from a place to B after 2.5 hours, and flies back against the wind for 3 hours. How many kilometers per hour does it fly back?

Let's fly x kilometers per hour when we return
According to a certain distance, speed and time are inversely proportional
x×3＝720×2.5
3x＝1800
x＝600

Change the quadratic function y = 1 / 2x ^ 2 + X + 2 / 3 to y = a (X-H) ^ 2 + K. thank you

Original formula = 1 / 2 (x ^ 2 + 2x + 4 / 3) = 1 / 2 (x ^ 2 + 2x + 1 + 1 / 3) = 1 / 2 [x - (- 1)] ^ 2 + 1 / 6
It can be reduced to y = 1 / 2 [x - (- 1)] ^ 2 + 1 / 6, where a = 1 / 2, H = - 1, k = 1 / 6

Finding the second derivative of y = xlnx, y = Xe ＾ x, y = x ＾ x
How can we find the derivative when there is an X in front and then an equation with x?
How do you get y '= LNX + 1?

There is a mistake on the second floor! My answer is correct! This question is mainly about the algorithm of derivative. That is, (UV) '= u'v + UV' (U, V are functions of x) by y = xlnx then y '= LNX + 1 then y' '= 1 / X by y = Xe ＾ x then y' = e ^ x + Xe ^ x then y '' = 2E ^ x + Xe ^ X by y = x ＾ x then y '= x ^ x (LNX + 1) then y' '= x ^ x (LNX + 1 / x + 1)

Addition and subtraction of fractions
（1）a-1-a²\a-1 （2）4\（a+2）+a-2
（3）（1-1\1-x）（1\x²-1） （4）m+2n\n-m + n\m-n - 2n\n-m

（1）-2
（2）a^2/(a+2)
3)x/(x-1)
4)1