If the square of 3m-2 of X and N-1 = 3 of 2Y are quadratic equations of X and y, then M + n =?

If the square of 3m-2 of X and N-1 = 3 of 2Y are quadratic equations of X and y, then M + n =?

(3m-2)^2=1
n-1=1
n=2
3m-2=1
m=1
3m-2=-1
m=1/3
m+n=3
m+n=7/3

It is known that the (imi-1) power of (m-2) x + the (n-square-8) power of (n + 3) y = 6 is a quadratic equation of two variables. The value of M and N can be obtained. If x = 2,1, the corresponding value of y can be obtained

Because it's a quadratic equation of two variables
So: imi-1 = 1. M = ± 2. M = - 2
(n square-8) = 1.n & sup2; = 9.n = ± 3.n = 3
So: M = - 2, n = 3
In this case, the equation is - 4x + 6y = 6
Substitute x = 1 / 2 to get
-2+6y=6
6y=8
y=4/3

Given that the equation x square a + 2 + 3Y square 1-2b = 15 is a bivariate linear equation about X and y, then the values of a and B are____

It is known that x ^ (a + 2) + 3Y ^ (1-2b) = 15 is a bivariate linear equation about X and y, that is, the exponents of X and y are all 1,
The solution is a = - 1, B = 0

Is the square of LG10 LG10 * LG10
Is the square of (LGA) LGA * LGA

(lg10)^2=lg10*lg10 =1
lg(10)^2=2

The operation LG10 + LG25 of the higher one logarithm

lg10+lg25 =1+lg25=1+1.398=2.398
lg10+lg25 =lg(10*25)=lg250
lg10+lg25=lg10+lg5^2=1+2lg5

It is known that the logarithm of 3 with 30 as the base is a, and the logarithm of 5 with 30 as the base is B. the value of logarithm of 3 + 5 with 30 as the base is? (expressed by A.B.)
Can you tell me the detailed process?

a+b=log(30)(3*5)=log(30)15
1-(a+b)=log(30)30-log(30)15=log(30)2
log(30)(3+5)=log(30)2^3=3log(30)2=3(1-a-b)

F (x) = Log1 / 3 ^ (x ^ 2 + 1) - log3 ^ (x + 3)!

f(x)=-log3^(x²+1)-log3^(x+3)
=-log3^[(x²+1)(x+3)]
f'(x)=-ln3(3x²+6x+1)/[(x²+1)(x+3)]
-3 < x < - 1 - √ 6 / 3
f'(x)

Let a = {x | 2 (Log1 / 2 ^ x) ^ 2-21 * log8 ^ x + 3 ≤ 0} if x ∈ a, the maximum value of F (x) = log2 ^ 2 ^ X / A * log2 X / 4 is 2
Make it clear. One hour

2(log1/2^x)^2-21*log8^x+3≤0,
2 [- log2 (x)] ² - 21 * [1 / 3log2 (x)] + 3 ≤ 0, that is, 2 [log2 (x)] ² - 7log2 (x) + 3 ≤ 0,
[log2 (x) - 3] [2log2 (x) - 1] ≤ 0, so 1 / 2 ≤ log2 (x) ≤ 3, that is √ 2 ≤ x ≤ 8,
So a = {x | √ 2 ≤ x ≤ 8}
The expression f (x) = log2 ^ 2 ^ X / A * log2 X / 4 is not clear. Is "x / a" the exponent of 2? Is "x / 4" true?
If f (x) = log2 (2) ^ (x / a) * log2 (x / 4) = (x / a) * log2 (x / 4),
When a > 0 and x = 8, the maximum value of F (x) is (8 / a) * log2 (8 / 4) = 2, that is, if 8 / a = 2, then a = 4;
When a

Solve the equation 10 ^ (x ^ 2) = 2 ^ (2x + 1). Log2 [2 ^ (- x) - 1] Log1 / 2 [2 ^ (- x + 1) - 2] + 2 = 0

1. Solve equation 10 ^ (X & # 178;) = 2 ^ (2x + 1)
Take the common logarithm on both sides to get: X & # 178; = (2x + 1) LG2, that is, X & # 178; - 2 (LG2) x-lg2 = 0,
So x = [2lg2 ± √ (4LG & # 178; 2 + 4lg2)] / 2 = LG2 ± √ (LG & # 178; 2 + LG2)
2.log₂[2^(-x)-1]log‹1/2›[2^(-x+1)-2]+2=0
The logarithm with (1 / 2) as the base is replaced by the logarithm with 2 as the base
-log₂[2^(-x)-1]log₂[2^(-x+1)-2]+2=0
That is to say, log &; [2 ^ (- x) - 1] Log &; {2 [2 ^ (- x) - 1]} - 2 = 0
log₂[2^(-x)-1]{1+log₂[2^(-x)-1]}-2=0
Let 2 ^ (- x) - 1 = u, then the above formula becomes (log &; U) (1 + Log &; U) - 2 = (log &; U) & # 178; + Log &; U-2 = (log &; U + 2) (log &; U-1) = 0
So from log &; U + 2 = 0, u &; = 1 / 4; from log &; U-1 = 0, u &; = 2
Then from 2 ^ (- x) - 1 = 1 / 4, we get 2 ^ (- x) = 5 / 4, - x = Log &; (5 / 4), so x &; = - Log &; (5 / 4) = 2-log &; 5;
From 2 ^ (- x) - 1 = 2, we get 2 ^ (- x) = 3, - x = Log &; 3, so x &; = - Log &; 3

Tan2 tan9 is bigger than the size

tan9=tan(9-2*3.14)>tan2