What is the smallest positive integer value of x where
y=2^x exceeds y=2x+1? State the corresponding value for y.
A. x=4,y=16
B.x=2,y=4
C.x=5,y=32
D. x=3,y=8
To find the smallest positive integer value of x where y=2^x exceeds y=2x+1, we can set the two expressions equal to each other and solve for x.
2^x = 2x + 1
To solve this equation, we can use trial and error with different values of x.
For x = 1: 2^1 = 2(1) + 1, which is false.
For x = 2: 2^2 = 2(2) + 1, which is true.
So, the smallest positive integer value of x where y=2^x exceeds y=2x+1 is x = 2.
To find the corresponding value for y, we can substitute this x value into either of the original equations.
Using y = 2^x, y = 2^2 = 4.
Therefore, the corresponding value for y is y = 4.
The correct answer is B. x = 2, y = 4.