It is known that the intersection coordinates of the first-order function Y1 equal to 3x + 3 and Y2 equal to 2x + 8 in the same rectangular coordinate system are (1,6), then when Y1 is greater than Y2, the value range of X is (1,6)

It is known that the intersection coordinates of the first-order function Y1 equal to 3x + 3 and Y2 equal to 2x + 8 in the same rectangular coordinate system are (1,6), then when Y1 is greater than Y2, the value range of X is (1,6)

y1=3x+3
y2=-2x+8
Suppose Y1 > Y2
It has 3x + 3 > - 2x + 8
Get 5x > 5
The solution is x > 1
That is, when x > 1, Y1 is greater than Y2

Compare the sizes of two exponential functions y = a ^ X. when a is greater than 1, but x is different, how to compare? For example, compare the sizes of Y1 = 4 ^ 0.8 and y2 = 8 ^ 0.48

Replace them all with exponential functions based on 2
It's 2 ^ 1.6 and 2 ^ 1.44
So Y1 is greater than Y2

Let Y1 = a ^ 3x + 1, y2 = a ^ - 2x, where a > 0, a is not equal to 1. When x is any value, there are (1) Y1 = Y2 (2) Y1 > Y2
exponential function

y1=y2
a^3x+1=a^-2x
3x+1=-2x
5x=-1
x=-1/5
y1>y2
a^3x+1>a^-2x
When a > 1
3x+1>-2x
5x>-1
x>-1/5
When 0