How to solve exponential function inequality (3 / 1) ^ x ^ 2-8 > 3 ^ - 2x?

How to solve exponential function inequality (3 / 1) ^ x ^ 2-8 > 3 ^ - 2x?


(1/3)^(x^2-8)>3^(-2x)
(1/3)^(x^2-8)>(1/3)^(2x)
The function y = (1 / 3) ^ x is monotonically decreasing on R
So there are:
x^2-8



Given that f (x) is an exponential function and f (- 3) = 8, find f (3)


F (x) is an exponential function
f(x)=a^x
f(-3)=a^(-3)=8
So f (3) = a ^ 3 = 1 / A ^ (- 3) = 1 / 8



Let y = y (x) be determined by the equation E ^ XY + cos (XY) = y and find Dy (0)
The answer is (1 + XY ') * e ^ y - (y + XY') = y '
When x = 0, y = 0, Dy (0) = DX
The answer is different


When x = 0, substituting into the equation, we get: 1 + 1 = y, we get: y = 2
The derivation of X is as follows
(y+xy')e^xy-sin(xy)*(y+xy')=y'
Substitute x = 0, y = 2 to get: 2 = y '
So Dy (0) = 2DX



Conversion of kilogram / cubic meter and gram / cubic centimeter


1kg / m3 = 1000g / cm3



Five nines add, subtract, multiply and divide equals 99


(9+9÷9)×9+9=99
In addition, please send and click my avatar to ask me for help,
Your adoption is the driving force of my service



Find the area of the triangle formed by the line y = 3x-1 and two coordinate axes


y=3x-1
The coordinates of the intersection point with the Y axis are (0, - 1)
The coordinate of the intersection point with X axis is (1 / 3,0)
Then:
The area of a triangle bounded by a line and two axes
=1/3*1*1/2
=1/6



If a cylinder with a diameter of 2 decimeters at the bottom is cut off and a cylinder with a height of 1 decimeter is cut off, the surface area of the original cylinder will be reduced______ Square decimeter


14 × 2 × 1 = 6.28 (square decimeter). A: the surface area of the original cylinder is reduced by 6.28 square decimeter



4.9.9.12 calculate 24 points


4*((9+9)-12)4*(9+(9-12))4*(9-(12-9))4*((9-12)+9)4*((9+9)-12)4*(9+(9-12))4*(9-(12-9))4*((9-12)+9)(4-(12/9))*9(4-(12/9))*99*(4-(12/9))((9+9)-12)*4(9+(9-12))*4(9-(12-9))*4((9-12)+9)*49*(4-(12/9))((9+9)-1...



Let f (x) have continuous derivatives, and f (x) = 0, f '(x) ≠ 0, f (x) = ∫ x (X & # 178; - T & # 178;) f (T) DT 0 (paired with X above)
When x → 0, f '(x) is infinitesimal of the same order as X & # 710; K=





How to convert square meter and meter?


No, the square meter is the square meter