A quantity Q increases at a rate proportional to its current size. If the initial population is given by Q0, then the expression for the time it takes for the population reach three times (treble) its initial size is Ln3/Q0.

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Step 1We will find out the required value. Step 2Let Q be the size of population at any time ' t 'then given that rate is directly proportionalto the size of population present at that time{:[" i.e. "(dQ)/(dt)prop Q],[=>Q^(')(t)=(dQ)/(dt)=kQ],[" Where "k=any" constant "],[=>(dQ)/(Q)=kdt]:}On integrating both sides, we get{:[Longrightarrow int(dQ)/(Q)=int kdt],[Longrightarrow ln(Q)-ln(N)=kt]:}Where " -ln(N) " is arbitrary constant{:[Longrightarrow ln((Q)/(N))=kt],[=>(Q)/(N)=e^(kt)],[=>Q(t)=Ne^(kt)]:}Given that, at t=0rarr Q=Q_(0)So from eqn. (1), we have{:[=>Q_(0)=Ne^(0)],[=>N ... See the full answer

Step 1We will find out the required value. Step 2Let Q be the size of population at any time ' t 'then given that rate is directly proportionalto the size of population present at that time{:[" i.e. "(dQ)/(dt)prop Q],[=>Q^(')(t)=(dQ)/(dt)=kQ],[" Where "k=any" constant "],[=>(dQ)/(Q)=kdt]:}On integrating both sides, we get{:[Longrightarrow int(dQ)/(Q)=int kdt],[Longrightarrow ln(Q)-ln(N)=kt]:}Where " -ln(N) " is arbitrary constant{:[Longrightarrow ln((Q)/(N))=kt],[=>(Q)/(N)=e^(kt)],[=>Q(t)=Ne^(kt)]:}Given that, at t=0rarr Q=Q_(0)So from eqn. (1), we have{:[=>Q_(0)=Ne^(0)],[=>N ... See the full answer