evaluating fcns; growth, exponential, and hyperbolic eqns

[fattbeezy]

1. On this one, I'm not sure how to set it up.
f(1) = 389
f(-1) = 11
for f(x) = 3 x 2^ax+b + 5

2. This one is a word problem and I can't find a growth formula:

City A presently has a population of 1000 and 2 years ago a population of 432. City B's population is now 750 and last year it was 465.

a) Rounded to the nearest year, when will their populations be equal?
b) Rounded to the nearest year, when will one city have a population double the other?

3. I'm not sure how to start this one.

Find all x which satisfy 7 * 4^x+1 -3*2^x+1=2^5

4. Find all x which satisfy 3coshx+2sinhx = 4

thank you

 

[stapel]

1) Your formatting is ambiguous: Do you mean "(3x)(2^ax) + b + 5", "3(2^ax + b + 5", "(3x)(2^{ax + b}) + 5", or something else? So we cannot provide specific hints, but, in general, plug the two given x-values into the function, and set equal to the given y-values. This gives you two equations in two variables. Solve the system.

2) What sort of growth is assumed? The exercise doesn't say and, with only two data points, any number of growth functions could be posited.

3) Your formatting is ambiguous, and the solution method will almost certain depend upon your meaning.

4) One way to get started might be to use one of the hyperbolic identities to convert the hyperbolic cosine into an hyperbolic sine. Then solve the resulting quadratic for the values of sinh(x). Then solve for the values of x.

If you get stuck, please reply showing all of your work, using standard formatting -- or else LaTeX -- to clarify your questions. Thank you.

Eliz.

 

[fattbeezy]

sorry my bad

f(x)=3(2^ax+b) + 5
f(1)=389
f(-1)=11
find f(0)

 

[stapel]

sorry my bad

f(x)=3(2^ax+b) + 5
f(1)=389
f(-1)=11
find f(0)

Okay... Now follow the instructions, provided earlier, to find the values of "a" and "b". Then evaluate at x = 0.

Eliz.