Suppose that the population of a colony of bacteria increases exponentially. If the population at the start is 300 and 4 hours later it is 1800, how long (from the start) will it take for the population to reach 3000?

A) #"4 ln10"/"ln6"# #"hrs"#

B) #"6/(4ln10)# #"hrs"#

C) #"(10ln4)/ln6# #"hrs"#

D) #"4(ln10-ln6)# #"hrs"#

Answer is A).

Thanks in advance!

A) #"4 ln10"/"ln6"# #"hrs"#

B) #"6/(4ln10)# #"hrs"#

C) #"(10ln4)/ln6# #"hrs"#

D) #"4(ln10-ln6)# #"hrs"#

Answer is A).

Thanks in advance!

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Jun 12, 2018

Answer:

See below.

Explanation:

We need get an equation of the form:

#A(t)=A(0)e^(kt)#

Where:

#A(t)# is the amounf after time t ( hours in this case ).

#A(0)# is the starting amount.

#k# is the growth/decay factor.

#t# is time.

We are given:

#A(0)=300#

#A(4)=1800# ie after 4 hours.

We need to find the growth/decay factor:

#1800=300e^(4k)#

Divide by 300:

#e^(4k)=6#

Taking natural logarithms of both sides:

#4k=ln(6)# ( #ln(e)=1# logarithm of the base is always 1 )

Divide by 4:

#k=ln(6)/4#

Time for population to reach 3000:

#3000=300e^((tln(6))/4)#

Divide by 300:

#e^((tln(6))/4)=10#

Taking logarithms of both sides:

#(tln(6))/4=ln(10)#

Multiply by 4:

#tln(6)=4ln(10)#

Divide by #ln(6)#

#t=color(blue)((4ln(10))/(ln(6)) " hrs"#

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