Log30 (90) is represented by log3 (2) = a, log5 (2) = B, a, B

Log30 (90) is represented by log3 (2) = a, log5 (2) = B, a, B

First of all, use the formula for changing the base. It's easy to write by changing it into natural logarithm. It's the same to change it into other numbers
Log3 (2) = LN2 / Ln3 = a obviously a and B are not equal to 0, so -- - > Ln3 = LN2 / a... 1)
log5(2)=ln2/ln5=b ----> ln5 = ln2/b ...2)
log30(90) = ln90/ln30 = ln(3*3*2*5)/ln(3*2*5) = (ln3+ln3+ln2+ln5)/(ln3+ln2+ln5) = 1+ln3/(ln3+ln2+ln5)
Replace Ln3 and LN5 by substituting 1 and 2, and reduce LN2 up and down to get the above formula = 1 + (1 / a) / (1 / A + 1 + 1 / b) = 1 + B / (a + B + AB)

Let LG2 = a, lg7 = B, use a, B to represent log8 9.8
Let LG2 = a, lg7 = B, denote log8 (9.8) with a and B
8 is the base

log8(9.8)
=lg9.8/lg8
=(lg98-lg10)/(3lg2)
=(lg49+lg2-1)/(3a)
=(2lg7+a-1)/(3a)
=(2b+a-1)/(3a)

It takes 8 hours for a truck to go from city a to city B, and 6 hours for a bus to go from city B to city A. the two cars are facing each other. How many hours later do they meet?

Meet in 3.5 hours

Definition: A is a rational number which is not 1. We call 1-1 / a the difference reciprocal of A. We know that A1 = 1 / 3, A2 is the difference reciprocal of A1, A3 is the difference reciprocal of A2 By analogy, A2012 =?

I'm a junior high school student. I don't know if it's right, but according to your definition, we can calculate A2 = 1 / (1 + 1 / 3) = 1 / 4 / 3 = 1 * 3 / 4 = 3 / 4, and so on, we can get A2 = 4, A3 = - 1 / 3, so we can see that the next cycle starts, that is, every two numbers have a cycle, so 2012 △ 2 = 1006, which can be said to be 1006 cycles and the remaining 0 numbers, Then the last value of a set of loops, that is, the value 4 of A2, is the value of A2012

(1 / 2) parachutists perform low altitude parachute jumping. When the helicopter is 224m above the ground, the athletes leave the plane to do free fall
(1 / 2) parachutists perform low altitude parachuting. When the helicopter is 224m above the ground, the parachutists leave the plane to do free fall. After a period of time, they open the parachute with an acceleration of 12.5m/s2

If the time is t, the velocity v = GT and the displacement H1 = GT ^ 2 / 2 when the parachute is opened
After opening the parachute, do deceleration movement, the acceleration is 12.5m/s ^ 2
Displacement h2 = V ^ 2 / 2A = (GT) ^ 2 / 2A
h1+h2=gt^2/2+(gt)^2/2a=L=224m
The solution is t = 5S

For a two digit number, the sum of one digit and ten digit number is 12. If one digit and ten digit number are exchanged, the new number will be 36 larger than the original number, and the original two digit number will be calculated

Let the number on the one digit be x, and the number on the ten digit be 12-x. the equation is 10 (12-x) + X + 36 = 10x + (12-x). The solution is x = 8, 12-8 = 4. Answer: the original two digit number is 48

Force analysis of front and rear wheels of automobile when they rotate

In the case of the front wheel drive, the friction on the front wheel is forward and the rear wheel is backward
In the case of rear drive, the friction on the front wheel is backward and the rear wheel is forward

Is the - 1 power of x plus 2 equal to 3 an equation of first degree with one variable? Our teacher said that it is

I don't think so. Here, the degree of the unknown x is - 1. The concept of unary linear equation: through simplification, the equation that contains only one unknown and the highest degree of the unknown is one is called unary linear equation. The usual form is ax + B = 0 (a, B are constants, and X ≠ 0). Unary linear equation belongs to integral equation

(1 / 2) two cars drive from a to B at the same time. The car travels 12 kilometers per hour more than the cart. The car arrives at B in 4.5 hours
(1 / 2) two cars drive from a to B at the same time. The car travels 12 kilometers per hour more than the cart. The car arrives at B in 4.5 hours. It returns along the original road and meets the cart 31.5 kilometers away from B

31.5÷（31.5×2÷12-4.5）
=31.5÷（5.25-4.5）
=31.5÷0.75
=42 km / h

2^2000/2^1999-2^2001
2^2000
_____________
2 ^ 1999-2 ^ 2001 is a simple calculation
Equivalent to 2 ^ 2000 (2 ^ 1999-2 ^ 2001)

2^2000÷(2^1999-2^2001)
=2^2000÷(2^2000*0.5-2^2000*2)
=2^2000÷2^2000(0.5-2)
=2^2000÷(-3\2*2^2000)
=2^2000÷(-3*2^1999)
=-2\3