a bacteria colony doubles in 7hrs. How long does it tae the colony to triple Use N=No2^t/T. where No is the intial number of bacteria and T is the time in hours it take the colony to double.
You just need
2^(t/7) = 3
(t/7) log2 = log3
t = 7log3/log2 = 11.1 hours
To find out how long it takes for the colony to triple, we can use the formula N = N0 * 2^(t/T), where N0 is the initial number of bacteria, T is the time it takes for the colony to double, and t is the time it takes for the colony to triple.
Given that the colony doubles in 7 hours, we know that T = 7 hours. Let's substitute this value into the formula:
N = N0 * 2^(t/7)
Now, we know that when the colony triples, it means that the final number of bacteria will be three times the initial number of bacteria, which can be expressed as N = 3 * N0. Substituting this into the formula, we get:
3 * N0 = N0 * 2^(t/7)
To solve for t, we need to isolate the variable. Let's divide both sides of the equation by N0:
3 = 2^(t/7)
Now, we need to solve for t. Since 2^(t/7) can be rewritten as 2^(t/7) = 2^log2(t/7), where log2(x) is the logarithm base 2, the equation becomes:
3 = 2^(log2(t/7))
To remove the exponent, we can take the logarithm base 2 of both sides:
log2(3) = log2(2^(log2(t/7)))
Using the property log2(a^b) = b * log2(a), the equation simplifies to:
log2(3) = (log2(t/7)) * log2(2)
Since log2(2) equals 1, the equation becomes:
log2(3) = log2(t/7)
Now we can isolate the variable by dividing both sides by log2(7):
log2(3) / log2(7) = log2(t/7) / log2(7)
Using the property logb(x) / logb(a) = loga(x), the left side simplifies to:
log7(3) = log2(t/7) / log2(7)
Finally, to solve for t, we can multiply both sides of the equation by log2(7):
t/7 = 7^(log7(3))
Multiply both sides by 7:
t = 7 * 7^(log7(3))
Using a calculator or logarithm table, you can now find the value of t.