Let x = (- 3) ^ 5 * (under the third root - 2) / (- 4) ^ 2 to find lgx

Let x = (- 3) ^ 5 * (under the third root - 2) / (- 4) ^ 2 to find lgx

First, simplify X
X=(3^5)*(2^(1/3))/2^4
lgX=lg((3^5)*(2^(1/3)))-lg(2^4)
=lg(3^5)+lg(2^(1/3))-lg(2^4)
=5lg3+(1/3)lg2-4lg2
=5lg3-(11/3)lg2

If 2lg (Χ - 3Y) = LG Χ + LG (4Y), then what is the value of Y / x?

A: LG (x-3y) ^ 2 = lg4xy
(x-3y)^2=4xy
x^2-6xy+9y^2=4xy
x^2-10xy+9y^2=0
(x-y)(x-9y)=0
x=y or x=9y
y/x=1 or y/x=1/9

A high school mathematics problem (about logarithm)
a^(lga-0.5)=√10
(lga-0.5) LGA = 0.5 can be obtained
I want to know how I got it
lg[a^(lga-0.5)]=lg(√10)
therefore
(lga-0.5)lga=lg[10^(1/2)]
How did you get it on the left?

a^(lga-0.5)=√10
Take common logarithm at the same time
lg[a^(lga-0.5)]=lg(√10)
therefore
(lga-0.5)lga=lg[10^(1/2)]
(lga-0.5)lga=0.5lg(10)
(lga-0.5)lga=0.5
Proof of the original formula

A rectangular playground, the perimeter is 150 meters, its length and width ratio is 3:2, the playground area is how many square meters?

Length: 150 △ 2 × 33 + 2 = 75 × 35 = 45 (meters), width: 150 △ 2 × 23 + 2 = 75 × 25 = 30 (meters), 45 × 30 = 1350 (square meters). Answer: the area of this playground is 1350 square meters

When Xiao Gang calculates the addition of two true fractions, he uses the rounding method to get the approximate value: 5 / 3 + 7 / 7 b = 1.38, and tries to find the value of a of B

1.38-3/5=0.78
1/7=0.142857…… 2/7=0.285714…… 3/7=4285714…… 4/7=5714285…… 5/7=714285……
A: the value is a / 5

A 470 meter long train passes the 1030 meter long bridge in 1 minute and 30 seconds, and then passes a tunnel in 45 seconds at the same speed. How long is the tunnel?

1 minute 30 seconds = 90 seconds, because the speed is the same, so according to the time of crossing the bridge and the time of crossing the tunnel,
The distance that the train passes through the bridge is 90 times that of the tunnel, so
The distance of the train passing through the tunnel should be (470 + 1030) △ 2 = 750 (m),
750-470 = 280 (m), so the tunnel length is 280 M

As shown in the figure, if the perimeter of square ABCD is 15cm, the perimeter of rectangular EFCG is 15cm______ ．

∵ quadrilateral ABCD is a square, the circumference of ABCD is 15cm, BC + CD = 7.5 (CM), EFCG is a rectangle, EFB = ∠ EGD = 90, the circumference of △ bef and △ DEG are isosceles right triangle, BF = EF, eg = DG, the circumference of EFCG is EF + FC

Let the integer a be expressed by scientific counting method to the 7th power of a × 10, then a is a few digits
also
If 860 million is expressed in the form of a × 10 to the nth power by scientific counting method, what is the value of N

Let integer a be expressed by scientific counting method to the 7th power of a × 10, then a is 8 digits
If 860 million is expressed in the form of a × 10 to the nth power by scientific counting, then the value of n is 8

The following two mobile phone billing tables consider the following issues
|Mode 1 | mode 2|
--------------|------------|------------|
Monthly rent | 30 yuan / month | 0|
--------------|------------|------------|
Local call fee | 0.30 yuan / min | 0.40 yuan / min|
① If the local call time is 200 minutes, how much is the charge for each of the two mobile phone billing methods?
② If the local call time is 350 minutes, how much is the charge for each of the two mobile phone billing methods?
③ For a local call time, when will the two billing methods charge the same amount?
④ If the call time is X minutes, how do you choose the billing method

1、 Mode 1 = 30 + 0.3x200 = 90
Mode 2 = 120
2、 Mode 1 = 135
Mode 2 = 140
3、 300 minutes
4、 Less than 300 minutes with mode 2 is greater than mode 1, which one 300 minutes at will

In isosceles trapezoid ABCD, ad is parallel, BC, m and N are the midpoint of AD and BC, e and F are the midpoint of BM and cm respectively
Find out if the quadrilateral menf is a diamond, if it is a square, explore the relationship between the height of trapezoid and the number of BC at the bottom

Because e and N are the midpoint of BM and BC respectively, en is the median line of triangle MBC, so en is parallel to MC. Similarly, it can be proved that FN is parallel to MB, so quadrilateral menf is parallelogram