Why log9 (9 / 2) = log3 (81 / 4)

Why log9 (9 / 2) = log3 (81 / 4)

Unequal
log3^2 (9/2)=1/2 log3 (9/2)
log3(9/2)^2=2 log3 (9/2)

Given log9 (8) = a, log3 (2) =? Is represented by A

First of all, with the formula of changing the bottom, log3 (8) / log3 (9) = a
So log3 (8) / 2 = a
So log3 (2 ^ 3) = 2A
So 3log3 (2) = 2A
So it's 2A / 3

It is known that log53 = a log54 = B
Prove log25 12 = 1 / 2 (a = b)

log5 3=a log5 4=b
5^a=3,5^b=4
5^(a+b)=3*4
[5^(a+b)]^2=12^2
25^(a+b)=12^2
[25^(a+b)]^(1/2)=(12^2)^(1/2)
25^[(a+b)/2]=12
So log25 12 = 1 / 2 (a + b)

most urgent
It is proved that there is no mapping from positive integers to positive integers, so that f (f (x)) = x + 2011, and there is f (f (x)) = x + 2010. If there is a wrong number before, this one will prevail

There are two ways for you
Method 1
Suppose there is such a function f, then f (x + 2011) = f (f (f (x))) = f (x) + 2011
By induction, f (x + 2011m) = f (x) + 2011m (M is an integer)
Let I, J belong to {0,1,2} 、2010｝=M
If f (I) ≡ J (MOD 2011), let J = f (I) + 2011k
Then f (J) = f [f (I) + 2011k] ≡ f [f (I)] ≡ I (mod2011)
Because the number of elements of M is odd, there is always x belonging to N, so that f (x) ≡ n (MOD 2011)
Let f (x) = x + 2011k (k is an integer)
Then f [f (x)] = f (x + 2011k) = f (x) + 2011k = x + 2011k * 2K (k belongs to integer)
Because f [f (x)] = x + 2011, so 2K = 1, contradiction
So it doesn't exist
Method 2 (this is relatively simple, but not easy to understand)
Suppose there is such a function f, then by using the known function equation, we can know that for any r belongs to {1,2,3 2010} = s, there exists that l belongs to s and R does not belong to s, such that f (R) = L + 2011, f (L) = R or F (R) = L, f (L) = R + 2011,
Define the mapping G: R → L (all belong to s), then l is bijective and if G (R) = L, G (L) = R,
This means that the number of elements in S is even and impossible

On the formula of logarithm function in high school mathematics

When a > 0 and a ≠ 1, M > 0, n > 0, then: (1) log (a) (MN) = log (a) (m) + log (a) (n); (2) log (a) (M / N) = log (a) (m) - log (a) (n); (2) log (a) (M / N) = log; (3) log (a) (m ^ n) = NLog (a) (m) (n ∈ R) (4) log (a ^ n) (m) = 1 / NLog (a) (m) (n ∈ R) (5) bottom changing formula: log (a) M = log (b) m / log (b) a (b > 0 and B ≠ 1) (6) a ^ (log (b) n) = n ^ (log (b) a) prove: let a = n ^ x, then a ^ (log (b) n = (n ^ x) ^ log (b) n = n ^ (x · log (b) n) = n ^ log (b) (n ^ x) = n ^ (log (b) a) (7) log identity: A ^ log (a) n = n; Log (a) m ^ (1 / N) = (1 / N) log (a) m ^ (- 1 / N) = (- 1 / N) log (a) M 2. Log (a) m ^ (M / N) = (M / N) log (a) m ^ (- M / N) = (- M / N) log (a) m 3. Log (a ^ n) m ^ n = log (a) M, Log (a ^ n) m ^ m = (M / N) log (a) M 4. Log (with a as the base under N root) (with m as the true number under N root) = log (a) m, log (with a as the base under N root) (with m as the true number under m root) = (n / M) log (a) M 5. Log (a) B × log (b) C × log (c) a = 1
The relationship between logarithm and exponent
When a > 0 and a ≠ 1, a ^ x = n, x = ㏒ (a) n

If the domain of function f (2x) is a closed interval from - 1 to 1, then the domain of function f (log2x) is,
(the first X is the square of 2, the second X is a true number)
When 0 ＜ x ＜ 1, the following equation holds,
A. (half) 1 + X ＞ (half) 1-x b.log (1 + x) (1-x) > 1
c. D. log (1-x) (1 + x) > 0
The sum of the maximum and minimum values of the function f (x) = ax (x is the square of a) + loga (x + 1) from 0 to 1 (closed interval) is a. then the value of a is

If x is between - 1 and 1, then the range of X of 2 is from half to 2, so the range of log2x is from half to 2, so the domain of definition is from two to four
As long as a is poor, B can judge that the base number is greater than one, while the true number is greater than zero and less than one, which is definitely wrong from the graph. Similarly, item D is not right, item C is 0 ＜ x ＜ 1, - 1 ＜ - x ＜ 0, 0 ＜ 1-x ＜ 1, and the square is also 0 to 1
When 0 ＜ a ＜ 1, ax (x is the square of a) and loga (x + 1) are decreasing functions, so the maximum is x = 0 and the minimum is x = 1. From the known equation, a is half. A ＞ 1, ax (x is the square of a) and loga (x + 1) are increasing functions, so the maximum is x = 1 and the minimum is x = 0. From the known equation, a is half
So it's half

Calculation of logarithm of grade one
lg21*lg50+lg25-lg51*lg20=
lg2*lg50+lg25-lg5*lg20=

Original formula = LG2 (lg5 + LG10) + lg25-lg5 (LG2 + LG10)
=lg2(lg5+1)+lg25-lg5(lg2+1)
=lg2lg5+lg2+lg25-lg2lg5-lg5
=lg2+lg25-lg5
=lg(2×25÷5)
=lg10
=1

The calculation of logarithm
1.lg14-2lg7/3-lg7-lg18
2.lg243/lg9
I'd better tell you how I got here

1.lg14-2lg7/3-lg7-lg18
=lg(2*7)-2(lg7-lg3)-lg7-lg(2*9)
=lg2+lg7-2lg7+2lg3-lg7-lg2-2lg3
=-2lg7
2.lg243/lg9
=lg(27*9)/lg9
=(5lg3)/(2lg3)
=5/2

High logarithm operation
Given log83 = P, log35 = q, then lg5 =?

By adding log83 = Lg3 / LG8 = lg3-lg8 = plog35 = lg5 / Lg3 = lg5-lg3 = q, we get: lg3-lg8 + lg5-lg3 = P + Q, that is: lg5-lg8 = P + Q8 = 1000 △ 125lg8 = lg1000-lg125 = 3-3lg5 substituting lg5-lg8 = P + qlg5 - (3-3lg5) = P + q4lg5-3 = P + qlg5 = (P + Q + 3) / 4

What is the monotone increasing interval of F (x) = Log1 / 3 (6-x-x ^ 2)

00
x²+x-6=(x+3)(x-2)