What is the growth rate of the E. coli population?

It's not mentioned in the question, but bacteria divide by binary fission, essentially doubling each time.

(A)

Use the formula:

\(A=A_0b^{\frac{t}{c}}\)

Let's fill this in:

\(123\ 456=57\left(2\right)^{\frac{6}{c}}\)

Solve for 'c', which represents the time it takes for the growth factor b to occur.

\(\ln \frac{123456}{57}=\ln \left(2\right)^{\frac{6}{c}}\)

\(\ln \frac{123456}{57}=\frac{6}{c}\cdot \ln \left(2\right)\)

\(\frac{\ln \left(\frac{123456}{57}\right)}{\ln \left(2\right)}=\frac{6}{c}\)

\(c=\frac{6}{\frac{\ln \left(\frac{123456}{57}\right)}{\ln \left(2\right)}}\)

\(c=\frac{6}{\frac{\ln \left(\frac{123456}{57}\right)}{\ln \left(2\right)}}=0.5414\)

(B) \(c=0.5414\ \rightarrow \ 5.41\times 10^{-1}\) bacteria / hour.

That's wrong

Well, nothing here mentions continuous or exponential growth, so I don't know.

Could be you subtract the smaller number by the smaller, then divide by 6. That'd suggest linear growth though, which makes little sense

try this

123 456 = 57 b ^ 6

123 456 / 57 = b ^ 6

log ( 123 456 / 57 ) = log ( b ^ 6 )

log ( 123 456 / 57 ) = 6 log b

log ( 123 456 / 57 ) / 6 = log b

10 ^ ( log ( 123 456 / 57 ) / 6 ) = b

b = 3.60 bacteria / hour

Each of the original 57 bacteria will divide 3.60 * 10 ^ 0 times per hour.

This is what I did:  

Growth rate formula: change in pop size divided by change in time


125,456 (final) - 57 (initial) / by 6 hrs

= 20899.83333 bacteria/hr

=2.08x10^4 bacteria/hr

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