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

1 answer

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.
Similar Questions
    1. answers icon 1 answer
  1. What is the smallest positive integer value of x where y=3xexceeds y=3x+3 ? What is the corresponding value for y?(1 point)
    1. answers icon 1 answer
  2. What is the smallest positive integer value of x where y=3xexceeds y=3x+3 ? What is the corresponding value for y?(1 point)
    1. answers icon 1 answer
  3. 1 of 51 of 5 ItemsQuestion What is the smallest positive integer value of x where y=3x exceeds y=3x+3 ? What is the
    1. answers icon 1 answer
more similar questions