Write the growth law

If bacterial growth is analogous to radioactive decay in form, then instead of decreasing exponentially, it increases exponentially: $N = N_0 e^{kt} \, , \quad k>0$

Take logarithm

Taking natural log on both sides: $\ln N = \ln N_0 + kt$

Rearranging in terms of the ratio: $\ln\left(\frac{N}{N_0}\right)=kt$

Interpret the graph

This means:

• $N$ vs $t$ is an exponential increasing curve.
• $\ln N$ vs $t$ is a straight line with positive slope.
• $\ln(N/N_0)$ vs $t$ is also a straight line through the origin.

Compare with radioactive decay

For decay: $N = N_0 e^{-\lambda t}$ so $\ln\left(\frac{N_0}{N}\right)=\lambda t$ is linear.

For growth: $N = N_0 e^{kt}$ so $\ln\left(\frac{N}{N_0}\right)=kt$ is linear.

Choose the correct option

Therefore, the correct graph must represent a straight line for $\ln(N/N_0)$ vs $t$ (or equivalently an exponentially increasing $N$ vs $t$). Among the given options, this corresponds to Option C.