Can you help me set up this problem please

P(t) = 100/1+48.2e^-0.52t

If the company's claim is true, how many months will it take to have mastered 80% of the program.

I just need help setting it up

Thanks!

I will assume you meant

P(t) = 100/(1+48.2e^(-.52t) )

You did not define P(t), but I will assume it will be the
"efficiency" mastered after t months
so .....

.8 = 100/(1 + 48.2e^(.52t))
.8 + 38.56e^(.52t) = 100
e^(.52t) = (100-.8)/38.56 = 2.572614
taking ln of both sides, and knowing that ln e = 1
.52t = ln (2.572614)
t = 1.817 or appr 1.8 months

check:
let t=1.817
P(1.1817) = 100/(1 + 48.2e^(.52(1.817)
= 0.8

Sure, I can help you set up the problem.

To find the number of months it will take to have mastered 80% of the program, we need to solve the equation P(t) = 0.8, where P(t) represents the progress of mastering the program over time t.

The given equation for P(t) is: P(t) = 100 / (1 + 48.2e^(-0.52t))

To set up the problem, we need to substitute P(t) with 0.8 in the equation, so it becomes:

0.8 = 100 / (1 + 48.2e^(-0.52t))

Now, to find the value of t, we need to solve for it. Here's one way to set it up:

1. First, let's multiply both sides of the equation by (1 + 48.2e^(-0.52t)) to get rid of the denominator:

0.8 * (1 + 48.2e^(-0.52t)) = 100

2. Next, distribute the 0.8 to both terms inside the parentheses:

0.8 + 38.56e^(-0.52t) = 100

3. Now, subtract 0.8 from both sides of the equation:

38.56e^(-0.52t) = 99.2

4. Finally, divide both sides of the equation by 38.56 to isolate e^(-0.52t):

e^(-0.52t) = 99.2 / 38.56

Now, you can take the natural logarithm of both sides to obtain t:

ln(e^(-0.52t)) = ln(99.2 / 38.56)

By taking the natural logarithm, we can isolate t and solve for it.