Anyone know the answer to this pre calculus question?

Question

A bacterial population of 6,000 doubles every 25 minutes, how long will it take for that population to reach 47,000. I cannot figure out how to make a formula or find the answer does anyone know, where to find out more information about exponential/logarithm word problems?

sry balinase your wrong...after 100 minutes the population is over 48,000

Answer 1

6000 times 2 to the x power equals 47000. solve for x. multiply your answer x by 25 minutes and you get the total number of minutes.

Answer 2

a million. Use the identity that cos 2x = 2cos² x - a million, and enable c = cos x for simplicity. Then: 2c² - a million - 2c = 2 2c² - 2c - 3 = 0 Use the quadratic formulation: c = (2 +- ?[(-2)² - 4(2)(-3)]) / (2*2) c = (2 +- ?[4 + 24]) / 4 c = (2 +- ?(4 * 7)) / 4 c = (2 +- 2*?7) / 4 c = (a million +- ?7) / 2 we may be able to no longer use the constructive root, as cos is under no circumstances more desirable than a million. So, remedy: cos x = (a million - ?7) / 2 2. (i am going to apply t for theta) change x = r cos t and y = r sin t. r² cos² t - 4r cos t + r² sin² t = a million (use sin² t = a million - cos² t) r² cos² t - 4r cos t + r² - r² cos² t = a million r² - 4r cos t - a million = 0 fixing for r utilising the quadratic equation again leads us to the reply. thus, it does no longer count number number even if we take the constructive or adverse root. both provide the same graph.

Answer 3

In 25 min there would be 12,000...50min..24,000 75min... 48,000. Is the progression linear? If so find the number per second and divide out to find the progression time to get to 47,000

Answer 4

A=P(e)^rt is the formula you need, I think.

To calculate for the time it would take the bacteria population to reach 47,000, equate the final amount (A) to the original amount(p) multiplied to e raised to the product of rate and time.

So, it becomes:

47000=6000*e^2t
47000/6000=e^2t

Take the natural log (ln) of both sides.

ln 47000/6000 = ln e^2t
I am assuming that you know that the ln e^x is just x. Therefore, ln e^2t is just 2t.

(ln(47000/6000))/2 = t.

Hope this helps. :)

Answer 5

6000*2^x = 47000

47/6 = 2^x

ln(47/6) = X*ln(2)
X = ln(47/6)/ln(2)
this will be how many 25 minute time periods will have transpired.

instead of X you may want to have X/25 to make it more clearly stated.... but most would agree it is self explanitory.

Answer 6

B(0) =6000
B(25) = 2B(0)
B(50) = 2B(1)
B(75) = 2B(2)
.
.
.
so it appears that:

B(3) = 2*B(2)=2*2*B(1)= 2*2*2*B(0)
B(75) = 2^3*6000

so the general equation is

B(n) = 6000*2^(n/25)

what is n when B(n) = 47000?

47000 = 6000*2^(n/25)

2^(n/25) = 7.83

take ln of both sides

(n/25) ln(2) = ln(7.83)

n = 25*ln(7.83)/ln(2)

n= 74.24 minutes

Answer 7

Ophello has it, and the answer is 74.24065877 minutes. or 74 minutes, 14.44 seconds (roughly)

Answer 8

Nope sorry

Answer 9

geometric sum
a1=6000
r=2
S_n=47000
n=?
n*25 minutes

Answer 10

Appx. 196mins.