Two pairs of logarithms are selected from 3 569 1015 to make their ratio equal to 1 / 3 Then choose two logarithms whose ratio is 1 / 2, and the composition ratio is () fast

3, 9, 5 and 15
3 and 6, 5 and 10

If the logarithm of 16 with a as the base is equal to 2, and the logarithm of 8 with four as the base is B, then a ^ B

It can be seen from the title
Loga16 = 2, then a = 4, if you change it into a power function, it will be a & # 178; = 16
You don't have to be second
a^b=4^(log4√8)=√8=2√2
This is a logarithmic identity

How to calculate 5.27 + 2.86-0.66 + 1.63 with simple operation (ladder equation)
Come on, it's today,

=（5.27+1.63）+（2.86-0.66）
=7+2.2
=9.2

27 × 3 + 25 × 4 △ 2 = 59, how to change one of the operation symbols to make the equation hold?

27/3+25×4÷2=59

If a and B are the two real roots of equation X05 + x-2011 = 0, what is the value of A05 + 2A + B?
What is the value of a & # 178; + 2A + B after X & # 178; + x-2011 = 0~

x²+x-2011=0 x²+x=2011 a+b=-1
a²+2a+b=a²+a+a+b=2011+(-1)=2011-1=2010

If both of the equations X & # 178; - 11x + 30-a = 0 are greater than 5, then the value range of the real number a is larger than 5

Let f (x) = x & # 178; - 11x + 30-a
If there are two equations, then △ > 0; the axis of symmetry is x = 5.5
If both of them are greater than 5, then there is
f(5)>0,△=11²-4(30-a)>0
The solution is 25-55 + 30-a > 0, A0, a > - 1 / 4
So the range of a is (- 1 / 4,0)

If both of the equations x ^ 2-11x + 30 + a = 0 are larger than 5, then the range of the real number a is larger than 5
Why is this discussion wrong

According to the meaning △ 0, X1 > 5, X2 > 5
Δ = (- 11) ^ 2-4 (30 + a) = 1-4a > = 0, a = X2, just calculate x2 > 5
The solution of the equation (11 - (1-4a) ^ (1 / 2)) / 2 > 5, a > 0 (A10, x1x2 > 25) can't guarantee X1 > 5, X2 > 5. It's like that human must be an animal, but animal is not necessarily human

It is known that the two equations X & # 178; - 11x + A + 30 = 0 of X are greater than 5, and the value range of a is obtained

Using the Veda theorem:
-b/a≥10
11≥10
accord with
c/a≥25
a+30≥25
a≥-5

Given that the two real roots of the equation x2-11x + m-2 = 0 are greater than 1, the value range of M is larger than 1

This mainly depends on the conditions to be guaranteed
First, (x1-1) (x2-1) > 0. This condition can ensure that X1 and X2 are greater than or less than 1 at the same time
Secondly, X1 + x2 > 0, x1x2 > 2, which can ensure that the two are positive
If x1x2 > 2, (x1-1) (x2-1) > 0, the case that both X1 and X2 are greater than 0 and less than 1 can be eliminated
Therefore, four conditions for determining the range of M are obtained
(x1-1)(x2-1)>0
x1+x2>0
x1x2>2
△≥0
The range of M can be obtained by using Veda's theorem

Proof: equation 2x2 + 3 (m-1) x + m2-4m-7 = 0, for any real number m, there are always two unequal real roots

For any real number m, (M + 7) 2 ≥ 0, the original equation has two unequal real roots. Therefore, for any real number m, the equation 2x2 + 3 (m-1) x + m2-4m-7 = 0 will always have two unequal real roots

Two pairs of logarithms are selected from 3 569 1015 to make their ratio equal to 1 / 3 Then choose two logarithms whose ratio is 1 / 2, and the composition ratio is () fast