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Solving an equation with the variable in an exponent usually involves use of logarithms. Although any base of logarithm can be used, it makes things simpler if you use the same base of logarithm as the base(s) that have an exponent with a variable. Since both bases with exponents are e, then we'll use base e (ln) logarithms:
The reason we use logarithms on problems like this is this property of logarithms:
. It allows one to take an exponent of the argument of a logarithm and move it in front of the logarithm. This is how we get the variable out of the exponent.
But to use this property, the exponent has to be the exponent of the entire argument. And our exponents apply to just the e's in the arguments and not to the 15 and the 13. Somehow we need to separate the 15 and the 13 from the e's. Fortunately there is another property of logarithms,
, which can be used to separate factors of an argument into separate logarithms. Using this on our two logarithms we get:
Now we can use the other property to move the exponents (of the logs that have exponents) out in front:
ln(15) + (0.08t)ln(e) = ln(13) + (0.09t)ln(e)
And since ln(e) = 1 by definition this simplifies to:
ln(15) + 0.08t = ln(13) + 0.09t
Now that the variable is out of the exponents, we can solve for it. Subtracting 0.08t from each side we get:
ln(15) = ln(13) + 0.01t
Subtracting ln(13) from each side:
ln(15) - ln(13) = 0.01t
The two logarithms on the left cannot be subtracted. They are not like terms. But we can use yet another property of logarithms,
, to combine them:
ln(15/13) = 0.01t
And finally we multiply both sides by 100:
100ln(15/13) = t
This is an exact expression for t.
P.S. When a problem refers to an exact solution it means "without a calculator" (because calculators use decimal approximations for more logarithms).