Log Application 1 – Compound Interest

Log Application 1 – Compound Interest

December 28, 2019 20 By Kailee Schamberger


>>In this video we do
an application problem that uses exponents
and or logs to solve. And so we do this
problem on this video.   Use the compound
interest formula to find how many years it
takes 1300 dollars invested at 9% compounded monthly to
increase to 2000 dollars. So what this means is if you
put 1300 dollars in a bank, and the interest rate is 9%
and it’s compounded monthly which means 12 times a
year, when is the value in that account at 2000 dollars? So what we want to do is
identify what we do know and use this formula here. So p stands for the original
investment, a is how much it is after a certain number of years. So let’s fill that in. We know that p is our original
investment of 1300 dollars and a, we want to find out when
it’s grown to 2000 dollars. All right, what about
the interest rate, r? Well that’s 9%, we want to write
that as a decimal or a fraction. So it’s .09 and we don’t
know the time, right. But n is how many times per
year the interest is compounded. Since it says monthly and there’s 12 months
in a year, n is 12. So we want to fill all these
numbers in to this formula and then we will compute
and find out what t is. So let’s do that. So we have a, what we’re
going to plug in 2000.   p is 1300 and we have 1
plus .09 divided by 12, we’d have r over n to the nt. So n is 12, so it’s 12t. So that’s the original
part of the problem. Now from here there’s
different ways you might, you know, simplify this. And I’m going to start by
simplifying inside parenthesis and at the same time dividing
both sides by 300, I mean 1300. So if I divide both sides
by 1300, okay I’m just going to leave that as 20/13. Okay, on the left side
20/13 reduces pretty easily. 20/13 equals, and now I have to
simplify inside this parenthesis because I have 1
in front of here so I could [inaudible]
1 as 12/12. Is everybody with me there? 12/12. So I have a
common denominator. So that means in a
fraction I would have 12.09 over 12 to the 12t. So this is what I have so far. Okay, at this point if you
can divide 12.09 divided by 12 and have it not being
a repeating decimal, then you can do that. If you get something that
12 doesn’t go into evenly, then what you would do
is just leave it as is. But in this case if you do
12, use your calculator, 12.09 divided by 12, you
get 1.0075, it does come out to an exact number
here to the 12t. But notice 20/13. If I try to divide that out,
it doesn’t come out exact. So if something doesn’t
come out exact, leave it in fractional form. All right so here’s the problem. We’re now going to try to solve
for t and t is in the exponent and the key is any time you
have a variable in the exponent that you’re trying to solve
for, you’re going to use logs. You’re going to take
the log of both sides or the natural log
of both sides. So we’re going to go on to the
next page to continue this. So this is what I have and
it’s up to you whether you want to use your log button or
your natural log button. I’m just going to use the
log button so I’m going to write the log of the left
hand side is equal to the log of the right hand side. Now if you want, you could
put brackets around here but it’s certainly not required. All right, so now if I have
this 12t, what I could do is since it’s in the
exponent, I could put it out in front of the log. So really it’s kind of
confusing seeing these brackets but I showed it just
that I’m saying that I’m doing it to both sides. So what do we have now? I’ve got this log
of 20/13, it’s okay, don’t worry about it,
just keep writing it. And we’re going to bring 12t out to the front times
the log of 1.0075. Now the trick here is you’re
trying to solve for t. So remember the log of 1.0075, that’s just some
number multiplied by t and then I also have 12 by t.
So what I’ve got here is the log of 20/13 and I’m going to
write this so you see this as a coefficient times
12 and I have log, 12 log of 1.0075 times t. So
what I want you to note is that this part here is
simply the coefficient of t. So we’re just going to divide
both sides by that number. So I’m going to divide the
right side and the left side. So I don’t do any rounding
until this very last step. And keep in mind this is
the log of 20/13 divided by 12 times the log of this. You can’t combine this
20/13 with this 1.0075, these are two separate
computations you’re going to have to make to solve
for t. Now if you put that in the calculator
with parenthesis, etcetera, so in the calculator I would
recommend you put like a bracket around the log here and then
parenthesis around the 20/13 and another bracket,
then a division sign and then another bracket,
12 times the log of this. You want to make sure that you’re very careful how you
enter this in your calculator. So go ahead and try
entering this in your calculator
and see what you get. It might take you a little
while to answer all that and this is what I
got in my calculator. So I’m going to put
approximately because now it’s not going to
be exact, it does go on and on. I got 4.8044, you know, and
it kind of goes on and on but it did say round to the
nearest tenth of the year. So I’m going to say that t
is approximately 4.8 years and in fact I’m going to, since
it’s a word problem it takes about 4.8 years, which
was the question. It says how long
did it take, right? How many years? Okay, wow, not an easy problem. You know at this point I would
suggest actually checking to see if this seems correct. So let’s go back to
the original problem. So here’s the original problem
and we already identified p, r, a, n, t but what I
want to do is not see if actually you invested this
1300 dollars at 9% at you know, compounded monthly
for 4.8 years, is it really 2000 dollars? So that’s what I
mean by checking. So I’m going to try that. So I’m going to say well
what is [inaudible]. And the question is, is it going
to come out to 2000 dollars, that what we would hope. So what would p be
without the 1300? I’m going to have 1 plus my
r is .09 divided by 12 again to the mt. So it’s 12 times,
now I’m going to actually plug in that number 4.8, okay. So we’re just going to
simply compute what happens if you put 1300 dollars
in for 4.8 years at this interest rate
compounded monthly. So again, we want to simplify
inside this parenthesis which was, I think 1.0075
and 12 times 4.8 is 57.6 so that’s how many times it
actually gets compounded. And now we’re going to use our
calculator to do the 1.0075 to the 57.6 and then
multiply it by 1300 or if you’re careful
you could go 1300 times, make sure there’s
parenthesis to do this. So go ahead now and
just see what happens when you simplify that. Be careful of your
order of operations. I usually go ahead of my
calculator just to make sure. I do this first and so I put
my calculator at 1.0075 raised at 57.6 that you do that
according to your calculator and then I press enter
and if you do that, you should have something
about 1.537854 etcetera. And then I just go ahead and
multiply by the 1300 next. And that gives me 1999.21, okay. Well that’s not exactly
2000 dollars right? We want to see if it’s
about 2000 dollars and of course it’s not going
to be because 4.8 was rounded. But it checks because that’s
pretty darn close, right. So we must have,
when we rounded, we rounded under a little bit
so that’s why when we plugged it in we got a little
bit smaller number. So now we verify that that does
make sense, about 4.8 years. So it takes about 4.8 years
was the answer to this problem.