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## Algebra 2 (Eureka Math/EngageNY)

### Unit 3: Lesson 10

Topic D: Lessons 23-24: Solving exponential models

# Exponential model word problem: bacteria growth

Sal evaluates an exponential function at a specific value in order to answer a question about an exponential model.

## Want to join the conversation?

• Do bacteria actually produce future generations exponentially like it is described in the problem?
• I think they do, pretty much like any other species. However, in case of bacteria cumulative outcome of this reproduction is very substantial and visible, as (under favourable conditions) they produce subsequent generations very frequently. Much more frequently than previous generations fade away.
• Wondering why this video is included in the Algebra II logarithms section. The problem doesn't require the use of logarithms.
• You're right that logs were not needed. This is because you were given a value for t. If the problem had asked you to find "t" when b(t) = 10,240, then you would have needed logarithms.
This is likely why the video is in the logarithm section.
• Shouldn't you divide by 120 for the last step?
• 120 is the amount of time given, so you don't divide by 120.
• At where did 2^5 come from?
• Since 2^10 is the same as 2^5 * 2^5 he was just breaking it down into something easier to calculate to verify that he had the right answer for 2^10 before working the rest of the equation.
• how to solve this question sir?
The number of cell double after each process of cell division every 2 hours. If there are 120 bacteria initially, how many bacteria will be there after one and half day?
(1 vote)
• If it doubles every 2 hours, you have a exponential function y=ab^x, a initial value b is base. So you have y=120 (2)^x. 1 day has 12 2 hour periods and 1/2 of a day has 6 two hour periods, so substitute for with x=18.
• What do you do if the e is the t?
• Suppose that a computer program is able to sort n input values in k x n^(1.5) microseconds. Observations show that it sorts a million values in half a second. Find the value of k.

May I ask how will you set up an equation for this?
(1 vote)
• The word problem states that the computer can sort n values in k * n^(1.5) microseconds. This gives us an equation of time to sort = k * n^(1.5). We are then given that it can sort one million values in half a second, this means our time would equal 500,000 microseconds (one million microseconds in a second). So we can then have an equation of 500,000 = k * (1,000,000)^1.5. From here the rest of the problem is simple algebra and you get your answer of k = 0.0005.
• 10^d/2 = 16,000

to

d/2 =log(16,000)

shouldn't it be division insted of multiplication ?

so d/2 = 16,000 / 10

= d/2 = log(1/10) (16,000) (maybe)

why? I'm confused, can anybody explain it(please) ?
(1 vote)
• log(16000) is a number that should be a little over 4 (think that log(10000) = 4). So I have no clue how you changed log(16000)=16000/10. So the opposite of dividing by 2 is multiply by which ends up with d=8.4 approximately.
• I came across this question which wasn’t in this video, but was wondering how to solve it.
ln√xx^k =2k
Note: ln has base of square root of x.

(1 vote)
• First - Some notation corrections. "ln" means "natural log" and always has a base of "e". You should be using log base √x.

Next, assuming your problem is: log√x (x^2) = 2k, this can't be solved. You have one equation and two variables. More info is needed to find the valueof either variable.

If you really are asking how does the expression: log√x (x^2) simplify to 2k, then here's how.
-- x is the same as having (√x)^2
-- [(√x)^2]^k can be simplified by multiply the exponents. This creates: (√x)^2k
-- So, your problem becomes: log√x (√x)^2k
-- The logarithm is asking you what exponent would you apply to √x to create (√x)^2k. The answer is 2k.

Thus: log√x (x^2) = log√x (√x)^2k = 2k

Hope this helps.