# Log Application 1 – Compound Interest

>>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.

Thanks, very helpful.

I really love your voice and the way you teach your subject. It helps me to understand logarithms and its application. tnx a lot.

Thank you very much ðŸ˜€

if APR 9 percent, how is that monthly compound interest becomes 9percent divided by 12?

It is wrong!

Super thanks a lot!

Thank you from 2016

Can someone please help me to calculate the days!!! I know how to do all that, however my question is asking for how many daysðŸ˜“

thanks Really Helped out!!

more helpful than the college "lecture" videos I had to pay for.

What would the N variable be if it was compounded yearly?

the numbers are big and i was scared but you said dont be scared so i wasnt scared and now i understand. i love you.

This was very helpful! Thank you!

Oh thank you so much. I was struggling with this things but I now understand.

Thank to do Much

thanks

Well done keepit up

THANK YOU IT HELPED LOADSSS

From 2019 ;)))

kindest math video ever

what if the problem is to find the number of compounding periods (n), anybody can help me solve it? thank you for the future responses.

so helpful!! thank you so much ðŸ™‚

thanks actually helped