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Answer: Collin: about $401 thousand Cameron: about $689 thousandStep-by-step explanation:A situation in which doubling time is constant is a situation that can be modeled by an exponential function. Here, you're given an exponential function, though you're not told what the variables mean. That function is ...[tex]P(t)=P_0(2^{t/d})[/tex]In this context, P0 is the initial salary, t is years, and d is the doubling time in years. The function gives P(t), the salary after t years. In this problem, the value of t we're concerned with is the difference between age 22 and age 65, that is, 43 years.In Collin's case, we have ... P0 = 55,000, t = 43, d = 15so his salary at retirement is ... P(43) = $55,000(2^(43/15)) ≈ $401,157.89In Cameron's case, we have ... P0 = 35,000, t = 43, d = 10so his salary at retirement is ... P(43) = $35,000(2^(43/10)) ≈ $689,440.87___Sometimes we like to see these equations in a form with "e" as the base of the exponential. That form is ...[tex]P(t)=P_{0}e^{kt}[/tex]If we compare this equation to the one above, we find the growth factors to be ... 2^(t/d) = e^(kt)Factoring out the exponent of t, we find ... (2^(1/d))^t = (e^k)^tThat is, ... 2^(1/d) = e^k . . . . . match the bases of the exponential terms (1/d)ln(2) = k . . . . . take the natural log of both sidesSo, in Collin's case, the equation for his salary growth is k = ln(2)/15 ≈ 0.046210 P(t) = 55,000e^(0.046210t)and in Cameron's case, ... k = ln(2)/10 ≈ 0.069315 P(t) = 35,000e^(0.069315t)