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## Algebra (all content)

### Course: Algebra (all content) > Unit 15

Lesson 4: Modeling with one-variable equations & inequalities# Exponential equation word problem

Sal models a context that concerns a bank savings account. The model turns out to be an exponential equation. Created by Sal Khan.

## Want to join the conversation?

- Why does sal say 1.2 after minute 1? I get him saying .2 as that comes from 20% but why 1.2?(17 votes)
- This is because he is just adding the 20 percent change to the original hundred percent to find out the answer. He is finding 20 percent of 6250 and then adding it to 6250 (which would be 100 percent) which becomes 120 percent or 1.2. This is the best explanation I can give, sorry if you didn’t understand. If you didn’t get this you should go and check out the previous units, where he explains it better.(2 votes)

- Why (1.2)^r and not (1.2)^(r-1) ?(5 votes)
- I'm assuming you mean "t" when you say "r". The reason it's not (1.2)^(t-1) is because 6250 is multiplied by 1.2 at the end of EACH year, starting with the FIRST.(5 votes)

- I tried doing this utilizing the geometric series equation (Sn = a(1-r^n)/1-r, but I don't seem to get the same answer(6 votes)
- why when you multiply the percent, why wouldn't it just be (.20)? Why add a one? That's whats just like throwing me off. Like where is this 1 coming from?(3 votes)
- If you multiply it by 0.2 you end up with 1/5th of what you began with. It's supposed to
*increase*. The "1" represents what you started with that year, and the additional 0.2 represents the 20%*increase*that year. Because 0.2 is 20% of one.(3 votes)

- How do you identify if you have an exponential sequence?(2 votes)
- You know you have an exponential sequence if the ratio between consecutive terms is a constant. What does this mean? Let's do an example:

Here is a sequence:

1, 2, 2, 4, 8, 32, 256

Is it exponential? Find the ratios

2/1 = 2

2/2 = 1

4/2 = 2

8/4 = 2

32/8 = 4

256/32 = 8

The ratios are all two for the first couple terms, but then increase to 4 and 8, so this can't be exponential.

Here's another

1, 1.1, 1.21, 1.331, 1.4641, 1.61051

Is it exponential? Find the ratios.

1.1/1 = 1.1

1.21/1.1 = 1.1

1.331/1.21 = 1.1

1.4641/1.331 = 1.1

1.61051/1.4641 = 1.1

All the ratios are the same, so the sequence must be exponential.

Basically, divide a term in a sequence by the previous term. If the quotient is always the same number, then you have an exponential sequence.

(Note: the quotient could be fractional and even negative. 1, -0.5, 0.25. -0.125, 0.0625. Here, the quotient is always -0.5)

Just for fun, can you tell me what the rule is for the first sequence I gave?(5 votes)

- Is this the fastest way:

6250(6/5)^t = 12960

(6^t)/(5^t) = 1296/625 = (6^4)/(5^4)

(6/5)^t = (6/5)^4

t=4(2 votes)- That could be the fastest way for many people, and it's extra nice because you don't necessarily need a calculator to solve it. Sal's way works best for me (since I use a calculator), but any way is fine if it gives you the correct answer.(3 votes)

- why does my calculator give log 12960 as 4.11261, whereas Sal calculates it as 2.0736? I know log to the base 10 would give us log 10,000 is 4.....(2 votes)
- The 2.0736 comes from 12960/6250. The total money in the account after t years can be modeled by A=6250*1.2^t. We want to know how long it will take for the account to reach 12960. Therefore, we plug in the 12960 into the equation: (12960=6250*1.2^t) After dividing and isolating t we are left with: (2.0736=1.2^t) This is where Sal uses the log function to solve for t.(2 votes)

- At7:24why does it not matter what base logarithm you use?(2 votes)
- Wrath,

I think it is because of the change of base property of logs. See this video:

https://www.khanacademy.org/math/algebra/logarithms-tutorial/logarithm_properties/v/change-of-base-formula-proof(2 votes)

- I need help on this question:

4^x-3*2^x+1+8<0(2 votes)- The starting equation is:

4^(x-3)*2^(x+1)+8<0

Since 4=2^2, we will add it in

2^(2)^(x-3)*2^(x+1)+8<0

We multiply the 2 and x-3 since it is a power to a power.

2^(2x-6)*2^(x+1)+8<0

Then we add 2x+6 and x+1.

2^(3x-5)+8<0

From here we subtract 8 from both sides

2^(3x-5)+8-8<0-8

2^(3x+5)<-8

If f(x)<g(x) then In(f(x))<In(g(x))

In(2^(3x+5))<In(-8)

Then we apply the log rule:

(3x+5)In(2)<In(-8)

I don't know what to do from here, so I just plugged it in a calculator, and I got...

No Solution.

And we are done.(2 votes)

- Hello everyone!

Could anyone explain how to solve this problem using the geometric sequence formula? If it's not applicable here could you give me a hint why?(2 votes)- Every year, the account balance is multiplied by 1.2 (we add 20%). So after n years, we've multiplied the balance by (1.2)ⁿ, so the balance is 6250(1.2)ⁿ. We can use the geometric sequence formula to evaluate (1.2)ⁿ, then multiply the result by 6250 to get a formula for the account balance after n years.(1 vote)

## Video transcript

Liam opened a savings
account and put $6,250 in it. Each year, the account
increases by 20%. How many years will it take
the account to reach $12,960? Write an equation that
models the situation. Use t to represent
the number of years since Liam opened the account. So I encourage you
to pause this video and actually try to do
it on your own first. Try to write this equation
that models the situation using the variable t in
the way they described. And then actually
answer the question, how many years will it take for
the account to reach $12,960? Well, let's just think about it. So t is to represent
the number of years since Liam opened the account. So let's just say it's
been 0 years since Liam opened the account. How much is he going to have? Well, he's just going
to have $6,250 in it. That's how much he starts with. Now, let's say it's
been one year since he opened the account. How much will he have? Well, he's going to
have $6,250 times-- or let's write it this
way-- plus 20% 6,250. It grows 20% every year. So this is how much he
started the year with, and then he gets another
20% of that 6,250. If we factor out
a 6,250, this is equal to 6,250 times 1 plus 20%,
or we could write that as 0.2. Which is equal to
6,250 times 1.2. Now, how much is he going to
have at the end of two years? Well, he's going to have the
same amount that he still had at the end of
one year, times 1.2. Because it grew by 20% again. So he's going to have
the amount that he had at the end of
one year times 1.2, which is equal to 6,250
times 1.2, times 1.2. Which is equal to 6,250
times 1.2 squared. I think you might see
where this is going. Or I could even write it like
this, order of operations, you do the exponent first. So what about after three years? So after three years, well,
we're just going to compound. We're going to multiply
by 1.2 once again. So then he's going to have 6,250
times 1.2 to the third power. And so after t
years, we're going to multiply by 1.2
that many times. So after t years in his account,
he's going to have 6,250 times 1.2 to the t'th power. 1.2 to the t'th power,
or to the t power. I don't want get
confused with the thing that you use to take bites with. Anyway. So they say, write an equation
that models the situation. So we want to figure out
how many years will it take the account
to reach 12,960? So we essentially
want to say, when is the account
going to be $12,960? Or we could write
12,960, when is that going to be equal to
6,250 times 1.2 to the t power? So that's the equation
right over there that models the situation. And then we need to think
about how we can actually go about solving this thing. Well, a natural thing is
to isolate the t variable. Let's divide both
sides by 6,250. So we could get-- and if
we flip the two sides, we could get 1.2 to the t power
is equal to-- well let me write this, 12,960 divided by 6,250. And since they're
both divisible by 10, why don't we divide
them both by 10? So it's 1,296 divided by 625. And there's several
ways that you could solve this
problem at this point. One way, if you feel
confident that this is going to have an integer
answer right over here, you could literally just
try to use your calculator and multiply 1.2 enough times
to get whatever number this is. And so we could do it that way. And as we'll see, there's
a more systematic way of doing it once you
learn about logarithms, and I'll do that at the end. But I'll do that
last just in case you haven't been exposed
to logarithms yet. So you could literally
say-- so let me just exit out of everything. So you could literally say OK,
let's see, 1,296 divided by 625 is this value. So let's see how many times
we have to multiply by 1.2. 1.2 times 1.2 gets
us-- well, that doesn't get us close enough. So let's try it three times. So let's take that same number. Let's just take
1.2, let's raise it. Let's raise 1.2. Let's just do it three times. Times 1.2 times 1.2. That still doesn't get us there. What if we were to multiply
by 1.2 one more time? Well, that actually
gets us there. And we just did
this by brute force. 1.2 to the fourth power
will give us this value. So that's one way, kind
of a brute-force way, of figuring out that
t is equal to 4. Another way, this just might be
a little bit less intuitive, it might jump out of
it, gee, this looks like some type of a power of 5. We know that 5 to the first
is 5, 5 squared is 25, 5 to the third is 125,
5 to the fourth is 625. And so you might recognize
this right over here is 5 to the fourth. And it's actually
a little bit harder to recognize that
this right over here is 6 to the fourth power. And this right over here is 6/5. So we could rewrite this as
6/5 to the t is equal to 6 to the fourth over
5 to the fourth. Which is the same thing as
6/5 to the fourth power. So here you'd say, well, 6/5 to
the t needs to be equal to 6/5 to the fourth power. t must be equal to 4. Now, this is nice
when you can recognize that this is something raised
to the fourth power, which isn't easy to do. Or if you know
this is an integer, and you can just keep
multiplying 1.2-- if you know it's a low integer. But the systematic
way of doing it is to actually use logarithms. And there's many
videos on Khan Academy about how to use logarithms. But if you're more
concerned with, well gee, if I just want to figure
out 1.2 to what power is equal to this thing,
what you would do-- and we prove this
in other videos-- is if you say, look, let's
take the thing that we want 1.2 to some power to be. Let's take the
logarithm of that. And actually, you could
take a logarithm any base. Your calculator tends to have
a natural log, which is base e, and a log base 10. We could just take
a log base 10. So let's do that. So we'll take the logarithm of
what we want to get to, 2.0736, and divide that by
the thing that we're trying to take the power
of to get to this number. So divided by the
logarithm of 1.2. And once again, we prove this--
actually, I wanted to divide. So let me insert
division symbol. So once again, it might
look like a little voodoo right here. We prove it in other
videos, but if you wanted to use a calculator to
calculate things like this, because sometimes it won't be
a nice integer number of years. It might be 3 and
1/2 years, or it might be 7.1234 years,
whatever it might be. This will give you a
more precise answer. So what do you want to get to? You want to get to 2.0736. What are you raising
to some power? 1.2. Divide the log of
the thing you're trying to get to divided
by the log of what the base that you're trying to raise to
a power, and you click Enter. And then you get--
so this is literally another way of saying that
1.2 to the fourth power is going to be 2.0736. So once again, if this
looks like voodoo, you don't know what
logarithms are, we have videos on
Khan Academy on that. But there's multiple
ways to tackle it, especially this problem
where the answer was a little bit simpler.