A radioactive nucleus $$A$$ with a half-life $$T$$, decays into a nucleus $$B$$. At $$t = 0$$, there is no nucleus $$B$$. At some time $$t$$, the ratio of the number of $$B$$ to that of $$A$$ is 0.3. Then, $$t$$ is given by: (Consider $$\log_{e} x = \log x$$)

Let the number of radioactive nuclei $$A$$ present initially (at $$t = 0$$) be $$N_{A0}$$. At any later time $$t$$, the number of nuclei $$A$$ left undecayed is governed by the exponential decay law

$$N_A(t) = N_{A0}\,e^{-\lambda t},$$

where $$\lambda$$ is the decay constant. First, we recall the relation between the half-life $$T$$ and the decay constant. By definition, after one half-life the number of nuclei is halved, so

$$N_A(T) = \dfrac{N_{A0}}{2} = N_{A0}\,e^{-\lambda T}.$$

Dividing both sides by $$N_{A0}$$ and then taking the natural logarithm, we obtain

$$\dfrac12 = e^{-\lambda T}\quad\Longrightarrow\quad -\lambda T = \ln\dfrac12 = -\ln 2,$$

so

$$\lambda = \dfrac{\ln 2}{T}.$$

Every decay of $$A$$ produces one nucleus of $$B$$ and we are told that $$B$$ itself is not radioactive. Therefore, the number of $$B$$ nuclei formed up to time $$t$$ equals the number of $$A$$ nuclei that have disappeared:

$$N_B(t) = N_{A0} - N_A(t) = N_{A0} - N_{A0}\,e^{-\lambda t} = N_{A0}\bigl(1 - e^{-\lambda t}\bigr).$$

We are given that at time $$t$$ the ratio of $$B$$ to $$A$$ nuclei is $$0.3$$, that is

$$\dfrac{N_B(t)}{N_A(t)} = 0.3.$$

Substituting the expressions for $$N_B(t)$$ and $$N_A(t)$$ found above, we have

$$\dfrac{N_{A0}\bigl(1 - e^{-\lambda t}\bigr)}{N_{A0}\,e^{-\lambda t}} = 0.3.$$

The factor $$N_{A0}$$ cancels, leaving

$$\dfrac{1 - e^{-\lambda t}}{e^{-\lambda t}} = 0.3.$$

We can rewrite the left side by bringing the denominator up:

$$\bigl(1 - e^{-\lambda t}\bigr) \cdot e^{\lambda t} = 0.3.$$

Distributing $$e^{\lambda t}$$ gives

$$e^{\lambda t} - 1 = 0.3.$$

Adding $$1$$ to both sides, we obtain

$$e^{\lambda t} = 1.3.$$

Now take the natural logarithm on both sides:

$$\lambda t = \ln 1.3.$$

Finally, substitute the earlier relation $$\lambda = \dfrac{\ln 2}{T}$$:

$$\dfrac{\ln 2}{T}\,t = \ln 1.3.$$

Solving for $$t$$, we get

$$t = T\,\dfrac{\ln 1.3}{\ln 2}.$$

Recognizing that the problem allows us to write natural logarithms simply as $$\log$$, the same result can be written as

$$t = T\,\dfrac{\log 1.3}{\log 2}.$$

Hence, the correct answer is Option C.