```Question 629399
A decrease of 6% each year means that one year's population will be 94% of the previous year's population. This makes the equation we should use:
{{{P = 13000*(0.94)^t}}}
We use 0.94, the decimal version of 94% since we don't use percents directly in equations. The "t" represents the number of years since 1995. And the "P" stands for the population t years after 1995.<br>
We're trying to find when the population reaches 5000. To find this we will replace the P with 5000, solve for t and the figure out what year that is:
{{{5000 = 13000*(0.94)^t}}}
TO solve for t we start by isolating the base and its exponent. Dividing both sides by 13000:
{{{5000/13000 = (0.94)^t}}}
which simplifies to:
{{{5/13 = (0.94)^t}}}
Next we find the logarithm of each side. Any base of logarithm can be used. But in this problem, since it seems we will want a decimal approximation of the answer, we will use a base your calculator "knows", base 10 or base e:
{{{ln(5/13) = ln((0.94)^t)}}}
Now we use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, which allows us to move an exponent in the argument out in front. (This is the reason we use logarithms to begin with. It is the normal way to move the exponent, where the variable is, down to where we "can get it" with regular algebra.)
{{{ln(5/13) = t*ln(0.94)}}}
Now we divide by ln(0.94):
{{{ln(5/13)/ln(0.94) = t}}}
This is an exact expression for the solution. For a decimal approximation we get out our calculators. (If your calculator has buttons for parentheses then you could just type the whole expression in at once.) You should get someting close to:
15.44250845 = t<br>
Since t represents the number of years since 1995, the year when the population will reach 5000 will be 1995+15.44250845 = 2010.44250845. So about halfway through 2010.<br.
Note: If you use base 10 logs, "log", instead of base e logs like we did above, you should get the same answer (even though the individual logs work out differently).```