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# Need help with this Logarithm Exponent Question?

So it's confusing me on how to solve this.

Bacteria have a doubling time of approx 10 hours. The starting population is 10,000 bacteria. How long will it take for there to be 100,000 bacteria?

### 3 Answers

- Anonymous1 month ago
hey china do you own damn covid calculations.

- Jeff AaronLv 71 month ago
b = 10000 * 2^(t/10) = 100000

2^(t/10) = 100000 /10000 = 10

Since t is a real number, then t/10 is a real number, so we have:

t/10 = log[2](10)

t = 10*log[2](10) =~ 33.219280948873623478703194294894 hours

That's 33 hours, 13 minutes, and around 9.411 seconds

- llafferLv 71 month ago
I use this general form for exponential growth/decay problems:

a(t) = ae^(kt)

Where:

a(t) is the value at time t.

a is the initial value

k is the growth/decay constant

t is the time since the initial measurement

You are told the double-life of this bacteria is 10 hours. So we can use this to solve for the growth constant by using these values:

a = 1

t = 10

a(10) = 2

Solve for the unknown k:

a(t) = ae^(kt)

2 = 1e^(k * 10)

2 = e^(10k)

ln(2) = 10k

ln(2) / 10 = k

If we put this into the general form we get:

a(t) = ae^(kt)

a(t) = ae^(t * ln(2) / 10)

We can now use this to answer the question. If we start with 10,000 bacteria (a) how long will it take for there to be 100,000 (a(t)). Substitute and solve for the remaining unknown (t):

100000 = 10000e^(t * ln(2) / 10)

10 = e^(t * ln(2) / 10)

ln(10) = t * ln(2) / 10

10 ln(10) / ln(2) = t

That's your exact answer. Using a calculator to get a decimal approximation we get:

t = 33.22 hours (rounded to 2 DP)