21) Two functions each have an initial value of 1. One function is linear and has a slope of 10 while the other function is exponential with a growth factor of 1.1.

A) At what x-value does the exponential function's output exceed the linear?

A) x = 68.55

Step-by-step explanation:

The initial value of a function can be the value of the function when x = 0.

So we have a linear equation:

f(x) = a*x + b

with a slope of 10, then:

f(x) = 10*x + b

and we know that:

f(0) = 1 = 10*0 + b

        1 = b

Then the linear function is:

f(x) = 10*x + 1

And we have an exponential function g(x)

A general expoential function is written as:

A*(r)^x

where r is the growth factor, here we know that it is equal to 1.1

Then:

g(x) = A*(1.1)^x

And we know that:

g(0) = 1, then:

A*(1.1)^0 = A*1 = 1

Then A = 1

g(x) = (1.1)^x

So our two functions are:

f(x) = 10*x + 1

g(x) = (1.1)^x

We want to find the value of x such that the exponential function's output exceeds the linear one. This is kinda hard to do analytically, so we can do it graphically.

Below you can see the graphs of both functions, where we can see that for x = 68.55, the exponential function's output exceeds the one of the linear function.

If we want to find the point analytically, we need to solve:

10*x + 1 = (1.1)^x

Which is a transcendental equation, which means that can not be solved in close form. So usually we solve them graphically.

Here we can also check that for the found value:

g(68.55) = (1.1)^68.55 = 687

f(68.55) = 10*68.55 + 1 = 685.6

So this is not the exact value at which g(x) overcomes f(x), but is a good approximation.

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