How to calculate log2 ^ x * log2 ^ 2x = 2

How to calculate log2 ^ x * log2 ^ 2x = 2

Let a = log2 ^ x
Then log2 ^ 2x = log2 ^ 2 + log2 ^ x = 1 + A
So a (a + 1) = 2
a²+a-2=0
a=-2,a=1
log2^x=-2,log2^x=1
So x = 1 / 4, x = 2

Given log2 (3) = x, find the value of (2 ^ (2x) - 2 ^ (- 2x)) / (2 ^ (x) - 2 ^ (- x))
No one can understand? Or did you not understand the title?

log2（3）=x
2^x=3
（2^(2x)-2^(-2x))/(2^(x)-2^(-x))
=[(2^x)^2-1/(2^x)^2]/[2^x-1/2^x]
=[(2^x+1/2^x)(2^x-1/2^x)]/[2^x-1/2^x]
=2^x+1/2^x
=3+1/3
=10/3

Find the maximum value of F (x) = log2 (x ^ 2-2x + 3)

Base 2 > 1, logarithmic function increases monotonically
f(x)=log2(x²-2x+3)=log2[(x-1)²+2]
When x = 1, X & # 178; - 2x + 3 has a minimum value of 2, then f (x) has a minimum value f (x) min = log2 (2) = 1
When X - > + ∞, X & # 178; - 2x + 3 - > + ∞, f (x) - > + ∞, has no maximum
The function has a minimum value of 1 and no maximum value

The minimum value of function y = log2 (x square minus 2x plus 5)

Let y = x ^ 2-2x + 5,
=(x-1)^2+4
Because y = log2 (x ^ 2-2x + 5),
Monotonically increasing on (0, + infinity),
So y = x ^ 2-2x + 5,
=(x-1)^2+4
When taking the minimum value, y = log2 (x ^ 2-2x + 5) is the minimum
So, when x = 1, y = log2, 4 = 2 is the minimum

Given that the range between the square of polynomial 2x + my-12 and the square of polynomial NX - 3Y + 6 does not contain x, y, find the value of M + N + Mn

The difference between the square of 2x + my-12 and the square of polynomial NX - 3Y + 6 does not contain x, y,
（2-n）x²+（m+3）y-18
therefore
2-n=0 m+3=0
n=2 m=-3
m+n+mn
=-3+2-3×2
=-7

It is known that the difference between the square of the polynomial 2x + my-12 and the square of the polynomial NX - 3Y + 6 does not contain x, Y. find the value of M + N + Mn
Just want to know one step, that is: if the two polynomials subtraction results do not contain unknowns
Then the coefficient subtraction of the unknown term in the two polynomials should be equal to 0
N-2 = 0, M - (- 3) = 0,
How do n and m work out? Let's be more careful in this step

2X ^ 2 - NX ^ 2 = 0 so 2-N = 0 so n = 2
Similarly, M = - 3

If the sum of the square of X + my-12y and the square of polynomial NX - 3Y + 6 does not contain x, y is used to find the value of Mn

x^2+my-12y+nx^2-3y+6=(n+1)x^2+(m-15)y+6
Without x, y, the coefficient of X, y is 0
n+1=0,m-15=0
n=-1,m=15

Given that the sum of polynomial 2x2 + my-12 and polynomial nx2-3y + 6 does not contain x, y, try to find the value of Mn

According to the meaning of the title: 2x2 + my-12 + nx2-3y + 6 = (n + 2) x2 + (M-3) y-6, ∵ and does not contain x, y, ∵ n + 2 = 0, M-3 = 0, that is, M = 3, n = - 2, then Mn = - 6

It is known that the polynomials 2x ^ 2 + my-12 + NX ^ 2-3y + 6 about X and y do not contain the term xy and try to find the value of Mn

So n + 2 = 0;
m-3=0;
So m = 3; n = - 2;
So Mn = - 6;
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Solution 2 / X-1 = 4 / X Square-1
Find the simplest common denominator first
Then take both sides

2 / X-1 = 4 / X Square-1
2 / X-1 = (2 / x) ^ 2-1 (using square difference)
2/x-1=(2/x-1)(2/x+1)
(2/x-1)(2/+1)-(2/x-1)=0
(2/x-1)(2/x)=0
2 / X-1 = 0 or 2 / x = 0 (no solution)
SO 2 / X-1 = 0 gives x = 2