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

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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

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Answer 1 · 2017-06-02T02:36:01

Answer: $t = \frac{3}{20} = 0.15$ $t=3/20=0.15$

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 $2^{10 t}$ $2^(10t)$ and the second by $4^{t} (8^{1 - 4 t})$ $4^t (8^(1-4t))$ , where T represents the time. Find $t$ $t$ when samples are equal.

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

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

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

Since we now have the same base on both sides, we can simply solve for $t$ $t$ 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:
${log}_{2} (2^{10 t}) = {log}_{2} (2^{2 t + 3 (1 - 4 t)})$ $log_2(2^(10t))=log_2(2^(2t+3(1-4t)))$

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

Now, we can solve for $t$ $t$ :
$10 t = 2 t + 3 - 12 t$ $10t=2t+3-12t$
$20 t = 3$ $20t=3$
$t = \frac{3}{20} = 0.15$ $t=3/20=0.15$

Therefore, at $t = \frac{3}{20} = 0.15$ $t=3/20=0.15$ 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 $2^{10 t}$ $2^(10t)$ :
$2^{10 \cdot \frac{3}{20}}$ $2^(10*3/20)$
$= 2^{\frac{3}{2}}$ $=2^(3/2)$
$= 2 \sqrt{2} \approx 2.82843$ $=2sqrt(2)~~2.82843$

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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

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