How solve it? In a laboratory a record of the number of bacteria is kept, in millions, that grow as a function of time for two different samples. If the first sample is expressed by 2^10t and the second by 4^t (8^1-4t), where T represents the time

1 Answer
Jun 2, 2017

Answer: t=3/20=0.15t=320=0.15

Explanation:

In a laboratory a record of the number of bacteria is kept, in millions, that grow as a function of time for two different samples. If the first sample is expressed by 2^(10t)210t and the second by 4^t (8^(1-4t))4t(814t), where T represents the time. Find tt when samples are equal.

To find tt where the samples have an equal amount of bacteria, we set the expressions equal to each other and solve for tt:
2^(10t)=4^t(8^(1-4t))210t=4t(814t)

Note that 44 can be written as 2^222 and 88 can be written as 2^323:
2^(10t)=2^(2t)(2^(3(1-4t)))210t=22t(23(14t))

Here, since we are multiplying two exponential terms with the same base, we can add the exponents:
2^(10t)=2^(2t+3(1-4t))210t=22t+3(14t)

Since we now have the same base on both sides, we can simply solve for tt in the exponents or a more formal way of explaining why we can do this is that we can take the log base 2 of both sides:
log_2(2^(10t))=log_2(2^(2t+3(1-4t)))log2(210t)=log2(22t+3(14t))

Since exponential and logarithmic functions are inverse functions, they cancel, giving us:
10t=2t+3(1-4t)10t=2t+3(14t)

Now, we can solve for tt:
10t=2t+3-12t10t=2t+312t
20t=320t=3
t=3/20=0.15t=320=0.15

Therefore, at t=3/20=0.15t=320=0.15 the bacteria samples have equal amounts.

If it is necessary to find the amount of bacteria in each sample, we can substitute into the first expression 2^(10t)210t:
2^(10*3/20)210320
=2^(3/2)=232
=2sqrt(2)~~2.82843=222.82843