# 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

Jun 2, 2017

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

#### 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 ${2}^{10 t}$ and the second by ${4}^{t} \left({8}^{1 - 4 t}\right)$, 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 $t$:
${2}^{10 t} = {4}^{t} \left({8}^{1 - 4 t}\right)$

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

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

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:
${\log}_{2} \left({2}^{10 t}\right) = {\log}_{2} \left({2}^{2 t + 3 \left(1 - 4 t\right)}\right)$

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

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

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