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Using A(t) = 100:
Now we solve for t. Since t is in an exponent, we start by isolating the base and its exponent. Dividing both sides by 500 we get:
Now we use logarithms. Any base of logarithm may be used. But certain bases usually make things easier:
A base of logarithm that matches the base of the exponent will make the expression for the exact solution simpler.
A base of logarithm that your calculator "knows" (base 10, log, or base e, ln) will result in an expression that is easy to convert to a decimal approximation (if one is wanted/needed).
In this case using base e logarithms will result in both of the above advantages. Using base e logs on each side:
Now we use a property of logarithms, , which allows us to move the exponent of the argument out in front of the log. (It is this property that is the very reason we use logs when solving equations where the variable is in an exponent. The property lets us move the exponent (where the variable is) out in front where we can "get at it" with algebra.)
By definition ln(e) = 1. (This is why matching the base of the log and the base of the exponent gives us a simpler expression.)
Last, we divide both sides by -0.1386:
This is an exact expression for the solution to the equation. If you want/need a decimal approximation use your calculator to find ln(0.2) and then divide it by -0.1386.
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