# Re: Solving for a variable in an exponent

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Posted by T.Gracken on November 06, 2002 at 10:04:24:

In Reply to: Solving for a variable in an exponent posted by Diana on November 06, 2002 at 00:55:23:

: The formula 5 * 2^(kt/.69) describes how an initial population of 5 billion will grow at a constant per capita rate of k persons per year per person.

: Use this formula to determine how many years t it will take for the population to double, to 10 billion.

: I know I can let y = 2^u and let u = kt/.69 and that the derivative is then 2^u (I did 2^u ln2* 1/.69). But I have no idea what to do next. Can anyone tell me how I am supposed to go about getting the kt down from exponent world? I have no idea how to isolate that variable.

first, you did not give us a formula. I will assume that you mean P=5 * 2^(kt/.69) is the formula where P is the population after t years.

also, without more information we can not get a decimal type result for t.

however, if the population doubles to 10 million after "t" years we would have:

10 = 5 * 2(kt/.69)

=> 2 = 2(kt/.69) ...divide both sides by 5

=> log(2) = log[2(kt/.69)] ...take log of both sides

=> log(2) = (kt/.69)*log(2) ...use property of logs: log[Nm] = m*log(N)

=> 1 = kt/.69 ...divide both sides by log(2)

=> 0.69/k = t ...multiply both sides by .69/k

the equation is now solved for t (in terms of k)

unless there is more information provided, I don't see how you can simplify the result further.

AND!!! what are you using derivatives for???

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