It is known that the second power of LG = a, the third power of LG = B. let a and B denote log 1 / 2

It is known that the second power of LG = a, the third power of LG = B. let a and B denote log 1 / 2

It is known that the second power of LG = a, the third power of LG = B. let a and B denote log 1 / 2
lg2=a lg3=b lg(15/2)=?right?
lg(15/2)=lg(30/4)=lg(3*10/4)=lg3+lg10-lg4=lg3+lg10-2lg2=b+1-2a=1+b-2a

How to write the process of negative 1 power of 2 × 64 power + LG2 × lg2500 + log power of 2 (lg5) - 4 + log (log (LG (lne))?

Log (LG (lne)) = 1 / 2 * 4 ^ (3 * 2 / 3) + LG2 * (LG25 * 100) + 2 (lg5) ^ 2-2 ^ (2 * log2 * 3) + log2 (log3 LG 1000) = 1 / 2 * 4 ^ 2 + LG2 * (LG25 + LG100) + 2 (lg5) ^ 2 - [2 ^ (log2 3)] ^ 2 + log2

(1) The solution set of LG (5x) * lgx = LG2 (2) log (x power of 4 + 144) - 4log 2 = 1 + log (x-1 power of 2 + 1) 5
The second question is based on 5
Ignore the following three "5"

1
lg(5X)*lgX=lg2
(lg5+lgx)*lgx-lg2=0
lg²x+lg5*lgx-lg2=0
lg²x-(lg2-1)lgx-lg2=0
∴(lgx+1)(lgx-lg2)=0
∴lgx=lg2,lgx=-1
∴x=2,x=1/10
two
log5 (4^x+144）-4log(5) 2=1+log (5)[2^(x-1)+1]
log5 (4^x+144）=log(5) 2^4+log(5) 5+log (5)[2^(x-1)+1]
log5 (4^x+144）=log(5){ 2^4*5*[(1/2)2^x+1]}
4^x+144= 80*[(1/2)2^x+1]
2^(2x)-40*2^x+64=0
2^x=t,
t²-40t+64=0,t=(40±√1344)/2 ===> x
(the root of this quadratic equation is so weird that it looks ugly.)
That's the way

Solving linear programming? Max z = X1 + 3x2, 5x1 + 10x ≤ 50, X1 + x2 ≥ 1, X2 ≤ 4, x1, X2 ≥ 0 by graphic method
Solving linear programming with graphic method?
max z = x1+3x2
5x1+10x≤50
X1+X2≥1
X2≤4
X1，X2≥0
It is pointed out that the problem has unique optimal solution, infinite solution, unbounded solution or no feasible solution?

As shown in the figure, the condition interval is the shadow on the way. The slope of Z = X1 + 3x2 = - 1 / 3, and Z is the ordinate of the intersection of the function and Y axis
It can be seen from the figure that when the function passes through point a, Z is the largest, the a coordinate obtained is (2,4), and z = 14 is obtained by substituting z = X1 + 3x2
So the maximum is 14
There is a unique solution

If the quadratic trinomial X & # 178; + 2mx + 2-m of X is a complete square, then we can find the value of M

X & # 178; + 2mx + 2-m is a complete square, which shows that X & # 178; + 2mx + 2-m = 0 has two identical real number solutions, so (2m) ^ 2-4 (2-m) = 0, the solution is m = 1 or M = - 2

As shown in the figure, De is the median line of △ ABC, M is the midpoint of De, and the extension line of CM intersects AB at point n, then s △ DMN: s quadrilateral anme equals ()
A. 1：5B. 1：4C. 2：5D. 2：7

∫ De is the median line of △ ABC, ∥ de ∥ BC, de = 12bc. If the area of △ ABC is 1, according to de ∥ BC, ∥ s ∥ ade = 14, am is connected. According to the title, s ∥ ADM = 12S ∥ ade = 18S ∥ ABC = 18, ∥ de ∥ BC, DM = 14bc, ∥ DN = 14bn, ∥ DN = 13bd = 13ad. ∥ s

What's the power of 2 to 1000? I don't want a decimal multiplied by 10 to the power of N, I just want an integer with several hundred digits

2^1000= 10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376
If you want to calculate the process, you can put forward

How to calculate the maximum power of a 220 V 15 a watt hour meter

p=220*15
=3300W
=3.3kW
That's it

It is known that M is an integer, and the equations 4x-3y = 6,6x + my = 26 have integer solutions
Finding the value of M

4x-3y=6 （1）
6x+My=26 （2）
To eliminate x, take the least common multiple 12 of 4 and 6,
Since 4 × 3 = 12 and 6 × 2 = 12, the formula (1) is × 3 and the formula (2) is × 2
12x-9y=18 (3)
12x+2My=52 (4)
(4) (3)
（2M+9）y=34
y=34/(2M+9)
Because x, y and m are integers, the value of 2m + 9 can be 2,17, - 2, - 17
After face examination, M can only be 4 and - 4

Questions about voltage, current, power and power generation efficiency
What is the relationship between voltage, current, power and power generation efficiency? Here is an example. Which of the two hollow cup speed reducers has higher power generation efficiency?
6V reduction motor
Working voltage: 6V
No load speed: 9020rpm
Speed after deceleration: 220rpm
Diameter: 16mm
Output power: 1W
No load current: 16mA
Locked rotor current: 420mA
Torque: unknown NM
Reduction ratio: 41:1
Reducer model: precision reducer (output ball bearing)
12V reduction motor
Working voltage: 12V
No load speed: 9500rpm
Speed after deceleration: 220rpm
Diameter: 22mm
Output power: 3.27w
No load current: unknown MA (output power is larger than the above 6V, this parameter must be larger than 6V, is that right?)
Locked rotor current: unknown MA (output power is larger than the above 6V, this parameter must be larger than 6V, is that right?)
Torque: 0.3nm
Reduction ratio: 43:1

If you use it to drive LED, I think it's still 12V. When the output voltage is 12V, the current output is 270ma. After reducing the voltage, you can basically drive LED