5. The bacteria population doubles every 10 hours. How long will it take to triple the initial amount? (Use log 3 / log 2 = 1.585)

==> I thought you could use the formula A(1/2)^(t/h), but I don't know... Any help is GREATLY appreciated!! :D

To solve this problem, we first need to understand the formula you mentioned: A(1/2)^(t/h). This formula is used to calculate the population of bacteria after a certain time, where A is the initial amount, t is the time elapsed, and h is the growth rate constant.

In this case, we want to find out how long it will take to triple the initial amount. So, let's assign some variables:

A = Initial amount of bacteria
t = Time in hours
h = Growth rate constant

According to the problem, the bacteria population doubles every 10 hours. This means that the growth rate constant (h) can be calculated as:

h = log 2 / 10

Now, we want to find the time it takes to triple the initial amount, which means the final amount (A_final) will be three times the initial amount (A):

A_final = 3A

Using the formula A(1/2)^(t/h), we can substitute A with 3A (since A_final = 3A) and solve for t:

3A = A(1/2)^(t/h)

Dividing both sides by A:

3 = (1/2)^(t/h)

Taking the logarithm (with base 1/2) of both sides:

log 3 = log [(1/2)^(t/h)]

Using the property of logarithms, we can rewrite this equation as:

log 3 = (t/h) * log (1/2)

Now, we can substitute the value of h:

log 3 = (t / (log 2 / 10)) * log (1/2)

Simplifying further:

log 3 = (t / (log 2 / 10)) * (log 1 - log 2)

Since log 1 = 0, the equation becomes:

log 3 = (t / (log 2 / 10)) * (-log 2)

Now, we can rearrange this equation to solve for t:

t = (log 3 / log 2) * (-log 2 / (log 2 / 10))

Simplifying further, using the given value of (log 3 / log 2 = 1.585):

t = 1.585 * (-log 2 / (log 2 / 10))

Dividing log 2 by log 2:

t = 1.585 * (-10)

Finally, calculating the value:

t = -15.85

Since time cannot be negative, we ignore the negative sign and conclude that it will take approximately 15.85 hours to triple the initial amount of bacteria.

To determine how long it will take to triple the initial amount of bacteria, we can use the formula for exponential growth:

N = N0 * (2^(t/10))

Where:
N = final population
N0 = initial population
t = time in hours

Since we want to triple the initial amount, N = 3N0. Substituting this value into the formula:

3N0 = N0 * (2^(t/10))

Dividing both sides by N0:

3 = 2^(t/10)

To solve for t, we can take the logarithm of both sides of the equation:

log 3 = log (2^(t/10))

Using the logarithmic property log a^b = blog a:

log 3 = (t/10) * log 2

Rearranging the equation:

t/10 = log 3 / log 2

Dividing both sides by 1/10 (or multiplying by 10):

t = (log 3 / log 2) * 10

Using the provided logarithmic value log 3 / log 2 = 1.585:

t = 1.585 * 10

Therefore, it will take approximately 15.85 hours to triple the initial amount of bacteria.