It is known that a and B are real numbers, and e < a < B, where e is the base of natural logarithm

It is known that a and B are real numbers, and e < a < B, where e is the base of natural logarithm

Proof: when e < a < B, to prove AB > Ba, as long as blna > AlNb is proved, that is, as long as LNAA > lnbb is proved, consider the function y = lnxx (0 < x < + ∞) because when x > e, y ′ = 1 − lnxx2 < 0, so the function y = lnxx is a decreasing function in (E, + ∞), because e < a < B, so LNAA > lnbb, that is ab > ba

It is known that a and B are real numbers, and e < a < B, where e is the base of natural logarithm

Proof: when e < a < B, to prove AB > Ba, as long as blna > AlNb is proved, that is, as long as LNAA > lnbb is proved, consider the function y = lnxx (0 < x < + ∞) because when x > e, y ′ = 1 − lnxx2 < 0, so the function y = lnxx is a decreasing function in (E, + ∞), because e < a < B, so LNAA > lnbb, that is ab > ba

It is known that a and B are real numbers, and b > a > C, where C is the base of logarithm of natural numbers. It is proved that B power of a > a power of B

First of all, B and a are greater than the base C of natural logarithm, then they are obviously greater than 1, and after taking logarithm, they are greater than 0. To prove a ^ b > b ^ A, as long as we prove blna > AlNb (taking natural logarithm on both sides), as long as we prove LNA / a > LNB / b. consider the function f (x) = LNX / x, find the derivative f '(x) = 1 / x ^ 2-lnx / x ^ 2 = (1-lnx) / x ^ 2, when LNX > 1, f' (x) a > C, so we have LNB > LNA > 1, and f (b) LNB / B, so we can prove the original proposition

B is greater than a is greater than e (natural logarithm base)
It's better to use derivative related knowledge

Certification:
Constructor f (x) = (LNX) / X (x > e)
∴f'(x)=(1-lnx)/x^2
∵x>e
∴lnx>1
∴f'(x)a>e
∴f(b)e>0
So AlNb

English Translation: no longer (phrase)

no more
no longer
not...any more
not...any longer

Given that x = 1 / 2 is the solution of the equation 2x + 3A = 4x-7, then a =?

2X + 3A = 4x-7, that is, 2x-3a-7 = 0
a=(2x-7)/3=-2

Summary of out phrases

To break out; cause to escape
Bring out To show; publish
To carry out; carry out (to the end)
Check out
Come out come out come out come out
Cut out
Drop out
Fall out
Figure out
Find out
To give out; emit (a smell, etc)
Go out
Hand out
Die out
Lay out, arrange, design
To omit; omit
Let out (water, etc)
Look out
Write, list, see, recognize
Pick out
Point out
To pull out; pull out
Put out put out put out put out
Run out (of)
To set out; set out
Stand out
Try out
Turn off (a radio, etc.); make; expel
Wear out
Wipe out
Work out
Out of Leave out of , lack of
Out of breath
Out of control
Out of date
Out of order
Out of place no longer in place; out of place
Out of practice
Out of sight
Out of the question
Out of question

Xiaojing arranged the natural numbers from 1 to 2009 in the following format: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 17 19 20 21

(first term + last term) × number of terms △ 2
（1＋2009）×2009÷2
=2010×2009÷2
=2019045

Nouns and adjectives of different countries

Big country:
Germany Germany
Japan
Korea Korea
American American
England English
France French
Canada Canada
Spain
Portugal
Russia
Italy
Sweden
Finland
Brazil
Arabian Arabian
Algeria, Algeria
Afghanistan
Argentina, Argentina
United Arab Emirates
Aruba
Oman
Azerbaijan, Azerbaijan
Egypt Egypt
Ethiopia
Ireland
Estonia, Estonia
Andorra, Andorra
Angola, Angola
Anguilla
Antigua and Barbuda, Antigua and Barbuda
Austria, Austria
Australia
Macau
Barbados, Barbados
Papua New Guinea, Papua New Guinea
Bahamas, Bahamas
Pakistan
Paraguay, Paraguay
Palestine, Palestine
Bahrain, Bahrain
Panama
Brazil
Belarus, Belarus
Bermuda
Bulgaria, Bulgaria
Northern Marianas
Palau, Belau
Benin, Benin
Belgium
Iceland, Iceland
Puerto Rico
Poland, Poland
Bolivia
Bosnia and Herzegovina
Botswana Botswana
Belize, Belize
Bhutan, Bhutan
Burkina Faso
Burundi Burundi
Bouvet Island
Korea, Democratic People's Republic of Korea
Equatorial Guinea
Denmark
Germany
East Timor, East Timor
Togo, Togo
Dominica Republic
Dominica
Russia
Ecuador
Eritrea, Eritrean
France
Faroe Islands
French Polynesia
French Guiana
French Southern Territories
Vatican
Philippines, Philippines
Fiji, Fiji
Finland
Cape Verde
Gambia, Gambia
Congo
Columbia, Colombia
Costa Rica
Grenada
Greenland, Greenland
Georgia
Cuba, Cuba
Guadeloupe
Guam
Guyana, Guyana
Kazakhstan, Kazakhstan
Haiti, Haiti
Korea, Republic of Korea
Netherlands
Netherlands Antilles
Head islands and McDonald Islands
Honduras, Honduras
Kiribati, Kiribati
Djibouti, Djibouti
Kyrgyzstan, Kyrgyzstan
Guinea, Guinea
Guinea pig Guine Bissau
Canada, Canada
Ghana

Given the function f (x) = LG (AX BX) (a > 1 > b > 0), if f (x) > 0 holds when x ∈ (1, + ∞), then ()
A.a-b>1
B.a-b≥1
C. If f (x) > 0 holds for X ∈ (1, + ∞), then ()

You can use exclusion
Because the function f (x) = LG (a ^ X-B ^ x) (a > 1 > b > 0) holds if x ∈ (1, + ∞)
If a-b0 does not hold
If A-B ≥ 1, then when x ∈ (1, + ∞), a ^ X-B ^ x is always greater than 1, and f (x) > 0 is constant
So choose B