An in depth examination of natural logs and exponentials.

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If you have e growing at a non-constant rate over a given interval, can you use the average rate to calculate the end growth?

ln(2)=1-1/2+1/3-1/4. This means that e^.69 is 2. This means that this is the exact number of growths (the rate at which a principle of 1 grows) that must occur in a given time for the principle to double.

Additionally, we can think of the natural log as what combination of time and rate are needed to get to x amount of my value. (, without whom, this intuitive understanding would have been much harder, has it listed as simply the time rather than a combination of the time and rate).

Betterexplained very astutely shows that the ln(.5) is a negative number because you’re trying to find out what combination of time and rate you need to get to half the amount you have now.

We can break the result of ln(.5) (which is -0.6931) into two components: rate and time.

We get the equation rate*time= -.6931. Now, we could have negative time, but it’s more plausible to have a negative rate.

If the two numbers in the in the log are inverse, then the result will be equal but of opposite sign. ln(3)= -ln(.33). This is because if you go from your amount to 3 times your amount, the growth rate*time will be equal to going from that new 3 times your amount to your original amount but in the opposite direction.

Now, what about the multiplication of logs?

That’s a hard question. But, I believe it can be answered. log represents the amount of growth (time*rate) that e (or 10) must go through before reaching a value. ln(10) is how long it will take for an original value to get to 10 times that value given continuos growth.

If you have 2ln(10), then you’re asking how long will it take for the original value to get to 100 times its current, or ln(10)+ln(10), which yields the same result: ten times original and then once again 10 times that number.

for ln(10)^2, would you be taking the ln(10) ln(10) times? Would that make sense? Would you need to do that? For some reason it feels like you could represent e^x^2 with that. That you could represent functions that are not elementary using that.