The meteorological observation data of the sounding balloon in a certain place shows that the temperature will decrease by about 6 ℃ for every 1 km increase in altitude. If the temperature in the region is 21 ℃ and the temperature in the upper air is minus 39 ℃, how many kilometers is the height obtained? There are some cards marked 6, 12, 18 and 24. The number on the last card is 6 larger than that on the previous card. Xiao Ming got three adjacent cards, and the sum of the numbers on these cards is 342 ① Which three cards did Xiao Ming get? ② Xiaohua said that he also got three adjacent cards, and the sum of the numbers on these cards is 86, do you believe it? We need to solve the equation Hurry!

Set the height to x km
21-6x = - 39, the solution is x = 10
Let the middle input be x, then the other two are X-6, x + 6
X + X-6 + X + 6 = 342 gives x = 114
The solution of x = 28.33 from the above method is obviously wrong, so it is not believed

The meteorological observation data of a sounding balloon show that the air temperature decreases about 6 ℃ when the height increases by 1 kilometer. If the ground temperature is 21 ℃ and the temperature in the upper air is - 39 ℃, how many kilometers is the height?

∵ when the height increases by 1km, the temperature decreases by about 6 ℃, the surface temperature of a certain area is 21 ℃, the temperature of a place in the upper air is - 39 ℃, and the height of this place is: (- 39-21) / (- 6) × L = 10 (km)

According to the experimental data of a certain place, the air temperature drops by 6 degrees when the height increases by 1km, and the known ground temperature is 21 degrees. (1) what is the temperature when the altitude is 8km? (2) the altitude when the altitude is - 24 degrees

(1) - 27 degrees, (2) 7.5km

The data observed by a meteorological detector in a certain place shows that the temperature will decrease by about 6 degrees for every 1 kilometer increase in altitude. If the ground temperature in this area is 21 degrees, the temperature measured at a place in the upper air is minus 9 degrees, and the altitude from the ground here is about () km?

（21+9）/6*1000=5000

3. The least common multiple of 12, 20, 30, 42, 56

3*2*2*5*7*2=840

Taking ㏒ 2 as the logarithm of the base 3 minus ㏒ 4 as the logarithm of the base 3 and ㏒ 6 as the logarithm of the base 3 is the specific process
The answer is 1. Oh, thank you

All the answers are in the picture

㏒ logarithm of base 35 with 5 - ㏒ logarithm of base 7 with 5 + ㏒ logarithm of base 7 with 5 - ㏒ logarithm of base 1.8 with 5

㏒ logarithm of base 35 with 5 - ㏒ logarithm of base 7 with 5 + ㏒ logarithm of base 7 with 5 - ㏒ logarithm of base 1.8 with 5
=Log is the logarithm with 5 as the base (35 ﹣ 7 / 3 × 7 ﹣ 9 / 5)
=Log base 5 (175 / 3)

(logarithm of base 125 with 2 + logarithm of base 25 with 4 + logarithm of base 5 with 8) (logarithm of base 2 with 5 + logarithm of base 4 with 25 + logarithm of base 8 with 125)

[(log2 125)+(log4 25)+(log8 5)][(log5 2)+(log25 4)+(log125 8)]=[(3log2 5)+(log2 5)+(1/3)(log2 5)][(log5 2)+(log5 2)+(log5 2)]=(13/3)(log2 5)(3)(log5 2)=13

Solving logarithmic equation 1) 8 (10 ^ - x) + 10 ^ x = 6 2) 10 ^ 3x-4 (10 ^ x) = 0 + please check the simplified problem
I've worked it out by myself, but I'm not sure
1）8(10^-x)+10^x= 6
2）10^3x-4(10^x)= 0
Simplify, please check
2/3 log x + 1/3 log xy^3
= log ( x^2/3) +log ( (xy^3)^1/3)
= log (x^2/3) + log log (xy)
= log (x^2/3) (xy)
= log x^5/3y

Let t = 10 ^ x (T > 0)
1) the original equation is changed to 8 / T + T = 6
T ^ 2-6t + 8 = 0 t = 2 or 4 10 ^ x = 2 or 4 x = log 2 or log 4
2) T ^ 3-4t = 0, t > 0, so t = 2, x = log 2
2/3 log x + 1/3 log xy^3
=2/3 log x + 1/3(log x +3log y)
=2/3 log x +1/3 log x +log y
=log x +log y
=log xy

Choose two teams from 2, 3, 4, 5, 6, 8, 10 and 12 to make their ratio equal and form a proportion. Then choose two logarithms whose ratio is 3,

12 6 8 4 12/6=8/4 (6 2 12 4)