The graph of an exponential function is given. What is the smallest positive integer value of x where this exponential function exceeds the linear function y=16x+1
? What is the corresponding y for the exponential function?
To find the smallest positive integer value of x where the exponential function exceeds the linear function, we need to set the two functions equal to each other and solve for x.
So, we have:
\(a^x = 16x + 1\)
In this case, the exponential function is represented as \(a^x\), where a is the base of the exponential function.
Let's assume a base of 2 for the exponential function. The equation becomes:
\(2^x = 16x + 1\)
To find the smallest positive integer value of x, we can start by substituting x = 1 and increase the value until the exponential function exceeds the linear function.
When x = 1:
\(2^1 = 16(1) + 1\)
\(2 = 16 + 1\)
\(2 = 17\)
Since 2 is less than 17, x = 1 is not the smallest positive integer value where the exponential function exceeds the linear function.
Let's try with x = 2:
\(2^2 = 16(2) + 1\)
\(4 = 32 + 1\)
\(4 = 33\)
Since 4 is less than 33, x = 2 is not the answer either.
Let's try with x = 3:
\(2^3 = 16(3) + 1\)
\(8 = 48 + 1\)
\(8 = 49\)
Since 8 is less than 49, x = 3 is not the answer.
Let's try with x = 4:
\(2^4 = 16(4) + 1\)
\(16 = 64 + 1\)
\(16 = 65\)
Finally, x = 4 is the smallest positive integer value where the exponential function exceeds the linear function. The corresponding y value for the exponential function can be found by substituting x = 4:
\(2^4 = 16\)
\(16 = 16\)
So, the smallest positive integer value of x where the exponential function exceeds the linear function is x = 4, and the corresponding y for the exponential function is 16.