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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?
• 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.
• Why (1.2)^r and not (1.2)^(r-1) ?
• 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.
• 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
• 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?
• 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.
• How do you identify if you have an exponential sequence?
• 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?
• 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
• 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.
• 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.....
• 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.
• At why does it not matter what base logarithm you use?
• I need help on this question:
4^x-3*2^x+1+8<0
• 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.