Treating the equalities $$\log_{10} 2 = 0.3010$$ and $$\log_{10} 3 = 0.4771$$ as exact, prove that the sum to 30 terms of the series $$2 + 6 + 18 + 54 + \dots$$ exceeds $$2 \times 10^{14}$$.

The series is geometric with first term 2 and common ratio 3, so the sum of 30 terms is $$\frac{2(3^{30} - 1)}{3 - 1} = 3^{30} - 1$$. Now $$\log_{10} 3^{30} = 30 \times 0.4771 = 14.313$$ and $$\log_{10}(2 \times 10^{14}) = 14.3010$$, so $$\frac{3^{30}}{2 \times 10^{14}} = 10^{0.012}$$. Since $$10^{u} > 1 + u$$ for $$u > 0$$, we get $$3^{30} > 2 \times 10^{14} \times 1.012 = 2 \times 10^{14} + 2.4 \times 10^{12}$$. Therefore $$3^{30} - 1 > 2 \times 10^{14}$$, as required.