Proving the constant rate of doubling for an exponential function

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SUMMARY

The discussion centers on proving that the time interval, d, for an exponential function Q=Pa^t remains constant when Q doubles. The user attempts to derive this by setting up the equation Q_0=Pa^t and Q_1=Pa^{t+d}, leading to the equation a^{t+d}/a^t = 2. The user contemplates using logarithms to simplify the equation, ultimately suggesting that d can be expressed as d = log(2^t) - t. The hint provided indicates a potential relationship between Q^{t+d} and Q^t plus Q^d, which may aid in the proof.

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Homework Statement


Given that Q=Pa^t and Q doubles between t and t+d, prove that d is the same for all t.

Homework Equations



Q=pa^t

The Attempt at a Solution



This is what I've tried so far:

Q_0=Pa^t and Q_1=Pa^{t+d}
Then:
\frac{a^{t+d}}{a^t}} \equiv 2

This is where I begin drawing blanks again. I want to say take the log, but I'm not sure if that is right.

If so, doesn't this give me:
\frac{t+d}{t} \equiv log(2) ?

Then from there: d \equiv log(2^t)-t

Would that be correct?
 
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Hint:
Q^{t+d}=Q^t+Q^d

Alternatively:
\log \frac{a}{b}= \log a - \log b
 

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