Step 1: Start from the exponential decay law instead of the log-ratio formula.
For a first order reaction, the concentration remaining after time $t$ follows $a - x = a\,e^{-kt}$, or in fraction form:
\[ \frac{a-x}{a} = e^{-kt} \]
Step 2: Fill in what "25% decomposed" means.
If 25% of the reactant has broken down, 75% remains, so $\dfrac{a-x}{a} = 0.75 = \dfrac{3}{4}$, and the time given is $t = 40$ minutes.
\[ \frac{3}{4} = e^{-k(40)} \]
Step 3: Take the natural log of both sides.
\[ \ln\left(\frac{3}{4}\right) = -40k \]
\[ -40k = \ln 3 - \ln 4 \]
Step 4: Convert natural log to base 10 log using $\ln x = 2.303 \log x$, so we can use the given log values.
\[ -40k = 2.303(\log 3 - \log 4) = 2.303(0.4771 - 0.6021) = 2.303 \times (-0.1250) \]
\[ -40k = -0.2879 \]
Step 5: Solve for k.
\[ k = \frac{0.2879}{40} = 0.0071975 \approx 7.19 \times 10^{-3}\ \text{min}^{-1} \]
Notice this reaches the same value as the log(a/(a-x)) route, just approached from the decay law side instead.
Step 6: Final answer.
\[ \boxed{7.19 \times 10^{-3}\ \text{min}^{-1}} \]