Given the function f (x) = log21 + x1 − x (I), prove: F (x1) + F (x2) = f (x1 + X21 + x1x2) (II) if f (a + B1 + AB) = 1, f (− b) = 12, find the value of f (a)

(1) This is a proof that: left = f (x1) + F (x2) = log21 + x1-1-x1-1 + log 21 + x1-1-x1 + 1-1-1 + log 21 (1 + x1-1-x1-1-1 + 1 + 1-1-x1-x2 = log2 (1 + x1-1-1-1-1 + 1-1-x1 + 21 + X1 + 1-x1 + 1-x1 + 21 + X1 + 21 + X1 + 21 + X1 + X1 + 21 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + X1 + x2 + x2 + x2 = log2 (1 + x1-1-1-1-1-1-1-1-1 + 1 + 1 + 1 + 1 + 1-x2-2-2-2-2-2 = log2 (log2 (1 (1 (1 (1 (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1+ AB), - 12 + F (a) = 1 The solution is f (a) = 32

Find the domain of the function f (x) = √ log (x + 2) with the base of 5

The domain must satisfy both:
(x + 2) ≥ 0 with log base 5
(x + 2) with log base 5 is greater than 1 with log base 5
x+2≥1
x≥-1①
X + 2 > 0, that is x > - 2
It is concluded that x ≥ - 1
So the domain is [- 1, + ∞)

We know that the function f (x) = log is based on 3, M
It is known that the definition field of logarithm of function f (x) = log with 3 as the base ((m (the square of x) + 8x + n) / the square of X + 1) is r, and the value field is [0,2]

Log3 (1) = 0, log3 (9) = 2. Let t = (MX ^ 2 + 8x + n) / (x ^ 2 + 1), then the problem becomes: the domain of definition is: (- ∞, + ∞), the range of value is [1,9], find the value of M, N. from t = (MX ^ 2 + 8x + n) / (x ^ 2 + 1), we get (M-T) x ^ 2 + 8x + (N-t) = 0, because x ∈ R, so the above quadratic equation about X has real roots, so Δ = 8 ^ 2-4 (...)

Ask, just inside, 2 log2 (3) power, how to change

Log2 (3) power of 2 = 3
General formula: loga (x) = x of a
Proof: loga (x) = loga (x)
Regarding the loga (x) on the right as a whole, we can get the loga (x) of a = x by transforming the original into an exponential expression

Find the minimum value of y = log (1 / 2) (x power of 1-3) - log2 (x power of 3 + one third)
(x power of 1-3) is true, half is base, (x power of 3 + one third) is true, and that 2 is base
It's LG (14 / 9),

y=log(1/2)(1-3^x)-log2(3^x+1/3)
=-log2 (1-3^x)-log2(3^x+1/3)
=-[log2(1-3^x)+log2(3^x+1/3)]
=-log2[(1-3^x)(3^x+1/3)]
Let t = 3 ^ x > 0
Then y = - log2 [(1-T) (T + 1 / 3)]
=-log2(-t²+2t/3+1/3)
=-log2[-(t-1/3)²+4/9]
≥-log2 (4/9)
If and only if t = 1 / 3, that is, 3 ^ x = 1 / 3, x = - 1, the minimum value - log2 (4 / 9) is obtained
So the minimum

Set a = {log2 (X-2) power ＜ 1} indicates that the set range is?

Because 02
(X-2) lglg22

What is the value range of X for log2 (- x) > x + 1?
Take 2 as the base and - x as the true number (left

By drawing, we can judge that f (x) = log2 (- x) and G (x) = x + 1 intersect at (- 1,0). The former is monotone decreasing, and the latter is monotone increasing. So only XX + 1 holds

If log2 (a) * log2 (b) = 1 (a > 1, b > 1), then the minimum value of AB is

Find the minimum value of log2 (a * b) first
log2(a*b)=log2(a)+log2(b)>=2sqrt(log2(a)*log2(b))=2
The minimum value of log2 (a * b) is 2
The minimum value of AB is 4
Note: sqrt is the root sign
If and only if a = b = 2, the minimum value is 4
In this problem, if a and B are symmetric, that is, a is replaced by B, B is replaced by a, and the formula remains unchanged. Generally, when a = B, the maximum value is taken

If log4 (3a + 4b) = log2 √ AB, then the minimum value of a + B is ()

There seems to be a lack of conditions: a > 0, b > 0
log4(3a+4b)＝log2√(ab)
↔log2(3a+4b)/2＝log2√(ab)
↔3a+4b＝ab
↔4/a+3/b＝1.
According to the basic inequality, it is obtained that:
a+b
＝(a+b)(4/a+3/b)
＝7+(3a/b)+(4b/a)
≥7+2√[(3a/b)·(4b/a)]
＝7+4√3.
When a = 4 + 2 √ 3, B = 3 + 2 √ 3,
The minimum value is 7 + 4 √ 3

Given that a, B ∈ R +, and log4 (2a + b) = log2ab, the minimum value of 8A + B is______ ．

∵ a, B ∈ R +, and log4 (2a + b) = log2ab = log4ab, ∵ 2A + B = AB, both sides are divided by AB, 1A + 2B = 1, ∵ a, B ∈ R +, ∵ 8A + B = (8a + b) (1a + 2b) = 8 + Ba + 16ab + 2 ≥ 10 + 2BA · 16ab = 18, if and only if Ba = 16ab, that is, a = 32, B = 6, 8A + B takes the minimum value 18

Given the function f (x) = log21 + x1 − x (I), prove: F (x1) + F (x2) = f (x1 + X21 + x1x2) (II) if f (a + B1 + AB) = 1, f (− b) = 12, find the value of f (a)