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

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. Find `t` when samples are equal.

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

Note that `4` can be written as `2^2` and `8` can be written as `2^3`:
`2^(10t)=2^(2t)(2^(3(1-4t)))`

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

Since we now have the same base on both sides, we can simply solve for `t` 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)))`

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

Now, we can solve for `t`:
`10t=2t+3-12t`
`20t=3`
`t=3/20=0.15`

Therefore, at `t=3/20=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)`:
`2^(10*3/20)`
`=2^(3/2)`
`=2sqrt(2)~~2.82843`