Algebra help needed!!
The population of bacteria in a petri dish triples every 10 minutes. The population of the bacteria is initially 400 organisms.

What unit is most appropriate for the time at which the number of bacteria is 600?

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Catya
The population triples ( P × 3 ) every ten minutes...

So if ten minutes is reached there will be 1,200 bacteria.

Edited - missed the question
** The units for time should be in minutes.

P = 400 × 3^(t/10)

at t = 0 ; P = 400
at t = 10 ; P = 1,200

When P = 600
600 = 400 × 3^(t/10)
6/4 = 3^(t/10)
Log(3/2) = Log(3^(t/10))
power rule
Log(3/2) = (t/10) Log(3)
10 × Log(3/2)/Log(3) = t

3.69 minutes the population will be at 600 bacteria
Unites for time are in minutes.

The population of bacteria in a petri dish triples every 10 minutes. The population of the bacteria is initially 400 organisms. The appropriate unit time at which the number of bacteria when it reaches 600 organisms is 3.690 minutes.

The population of bacteria in a Petri-dish triples every 10 minutes.

It implies that the population of bacteria can be algebraically represented as:

= ( y × 3)

Now, after 10 mins, for 400 organisms, the population of the bacteria will be:

= 400 × 3

= 1200

Thus,

  • at the initial time t = 0 , there are 400 organisms.
  • At time t(10), there are 1200 organisms

The expression used to estimate the unit time for a population of  600 bacteria can be computed as:

[tex]\mathbf{y = 400 \times 3^{(\frac{t}{10})}}[/tex]

where,

  • y = number of bacteria

[tex]\mathbf{600 = 400 \times 3^{(\frac{t}{10})}}[/tex]

[tex]\mathbf{\dfrac{600}{400} = 3^{(\frac{t}{10})}}[/tex]

[tex]\mathbf{\dfrac{3}{2} = 3^{(\frac{t}{10})}}[/tex]

[tex]\mathbf{log (1.5) = log (3^{(\frac{t}{10})}})[/tex]

Using power rule

[tex]\mathbf{log (1.5) =\dfrac{t}{10} log (3)}[/tex]

t = 10 × 0.3690

t = 3.690 minutes

Therefore, we can conclude that the unit time most appropriate for the number of bacteria of 600 organism is 3.690 minutes.  

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