How to deduce the deduction of logarithm exchange formula? Inference~

How to deduce the deduction of logarithm exchange formula? Inference~

Two fractions with different denominators cannot be added directly. They can only be added after changing into the same denominator. In the same way, logarithms with different bases need to be changed into the same base if they want to operate with each other. In this way, the formula for changing the base is produced ① Then B = Logan ② The logarithm identity is obtained by substituting 2 into 1
3x^3-3x^2-4x=24
X = 2.646267 the problem is wrong... I solved it by programming
4.6x-2.4x is equal to? (lie equation)
Please pull~```
I need it urgently```
Help~`````
I can only say that it's 2.2x
A freight car is running on the railway at the speed of 8 m / s. due to the dispatching accident, an express car is running on the same track at the speed of 20 m / s at 600 m behind in the fog. At this time, the driver of the express car receives a report that a freight car is running in front of him. The driver of the express car quickly closes the brake, but the express car has to slide for 2000 m before stopping. Please judge whether the two cars will collide
The acceleration of the fast train after braking is & nbsp; & nbsp; a = V2 − v202x = 0 − 2022 × 2000m / S2 = - 0.1m/s2. When the two speeds are equal & nbsp; t = V − v0a = 8 − 20 − 0.1s = 120s, the forward distance of the fast train is x fast = v0t + 12at2 = 20 × 120 + 12 × (- 0.1) × 1202m = 1680m, and the forward distance of the slow train is x slow = V slow t
What is the maximum value of 4x-3x ^ 2-8?
y=3/20,x=2/3
20/3
x=2/3
y=3/20
How much is 6x - (4x-3) = 17? Write down the solution
6x-4x+3 = 17
2X=14
X=7
6x-(4x-3) = 17
6x-4x+3=17
6x-4x=14
2x=14
X=7
6x-4x=17-3,2x=14,x=7
Just try:
6x-(4x-3)=17
Get rid of the brackets
6x-4x+3=17
2x+3=17
Subtract 3 from both sides at the same time
2x+3-3=17-3
2x=14
X=7
Physics problem, how to solve the formula steps?
When the ball falls freely, the diameter of the ball is 0.5cm, the reading time from the photoelectric gate timer is 0.001s, and G is taken as 10m / s, the distance between the falling position of the ball and the photoelectric gate
V = D is greater than t, and the diameter and switch time are brought in
T = V to g, t is the falling time
X = - 189; gt ^ 2, find the distance
1/2 g t1^2 =h
1/2 g t2^2 =h+0.0005
t2-t1 = 0.001
Three equations and three unknowns are solved.
Although the title is not clear, I'd better solve it according to the conjecture
I understand that the time on the timer is the time when the ball passes through the photoelectric gate
The key to this problem is to find the final velocity V of the ball
V instantaneous = V initial + GT1 T1 is the initial velocity of the ball passing through the photoelectric gate
D diameter = V initial T1 + 1 / 2gt1 * T1
The initial value of V is obtained,
Because of the free falling body, then H = V initial * V initial
Although the title is not clear, I'd better solve it according to the conjecture
I understand that the time on the timer is the time when the ball passes through the photoelectric gate
The key problem is to find the minimum velocity of the ball
V instantaneous = V initial + GT1 T1 is the initial velocity of the ball passing through the photoelectric gate
D diameter = V initial T1 + 1 / 2gt1 * T1
The initial value of V is obtained,
For a free falling body, then H = vinitial * vinitial / 2G
The drop distance of the ball is
H = H + 1 / 2D
(x-3) & sup2; = (3x-2) & sup2; the equation is solved by factorization
（x-3）²=（3x-2）²
The equation is solved by factorization
(3x-2)²-(x-3)²=0
Square difference
(3x-2+x-3)(3x-2-x+3)=0
(4x-5)(2x+1)=0
x=5/4,x=-1/2
x²-6x+9=9x²-12x+4
-8x²+6x+5=0
△=b²-4ac=36+160=192
x1= x2=
x1=-4/5 x2=5/2
How to solve the equation 6 (X-5) = 4x?
6(x-5)=4x
How to solve this equation? It must be complete, not less than one step! Finally, X / 2 = 30 / 2 must be calculated!
6(x-5)=4x
6x-6x5=4x
6x-30=4x
6x-4x=30
(6-4)x=30
2x=30
x=30÷2
x=15
6x-30=4x 6x-4x=30 2x=30 x=15
6(x-5)=4x
6x-6×5=4x
6x-4x=6×5
(6-4)x=30
2x=30
2x÷2=30÷2
x=15
6(X-5)=4X
6X-30=4X
2X=30
X=15
6(x-5)=4x
6x-30=4x
6x-4x=30
2x=30
x=30/2=15
6x-30=4x
30=6x-4x
30=2x
x=15
6(x-5)=4x
6x-6*5=4x
6x-30=4x
6x-4x=30
2x=30
x=30/2
=15
6X-6×5=4X
6X-30=4X
6X-4X=30
2X=30
X=30÷2
X=15
First question
6(x-5)=4x
6x-30=4x
10x=30
X=3
Second question
x/2=30/2
2X=2x30
x=30
6x-30=4x
6x-4x=30
(6-4)x=30
2x=30
x=30/2
x=15
On the horizontal plane of smooth insulation, two point charges a and B with positive charge are fixed, the distance between them is r, QA = 2qb = q, and the point charge C with static released mass m and positive charge q is released at the midpoint of AB line. The magnitude and direction of acceleration of C when it is just released and the position when C has the maximum velocity are calculated
When C is just released: FA = k * QA * q / (R / 2) ^ 2 between a and C; FB = k * QB * q / (R / 2) ^ 2 between B and C; f (∑) = fa - FB; a = f (∑) / M = 2 * Q * q / R ^ 2 direction points to the maximum velocity of B, the resultant force of C is zero: FA = FBK * Q * q / La ^ 2 = k * Q * q / 2 * L
The acceleration is calculated by Coulomb's law, and then the force is synthesized. When the acceleration is zero, the maximum velocity is calculated by the positive ratio of the amount of charge and the square of the distance.
The first question is to find out the force of a and B on C (using the formula F = kqq / R ^ 2). And then the acceleration.
Second question. Obviously, the force of a on C is large, so C moves to B. In this process, the force of a on C decreases, the force of B on C increases, and the acceleration a decreases. When a is zero, the force of a and B on C is equal, and the velocity is the maximum. That is to find out the position where a and B have the same force on C. Let AC distance be l, then BC distance be r-l. Then use the formula set in the first question to find L. ... unfold
The first question is to find out the force of a and B on C (using the formula F = kqq / R ^ 2). And then the acceleration.
Second question. Obviously, the force of a on C is large, so C moves to B. In this process, the force of a on C decreases, the force of B on C increases, and the acceleration a decreases. When a is zero, the force of a and B on C is equal, and the velocity is the maximum. That is to find out the position where a and B have the same force on C. Let AC distance be l, then BC distance be r-l. Then use the formula set in the first question to find L. Put it away