Q.Given $\log 2 = 0.3010$ and $\log 3 = 0.4771$ (base 10), find

Each number is rebuilt from 2, 3 and 10 so the product, quotient and power laws can be applied to the two given logarithms.

Using $\log 2=0.3010$, $\log 3=0.4771$ and $\log 10=1$: $\log 6 = 0.7781$, $\log 12 = 1.0791$, $\log 5 = 0.6990$ and $\log 1.5 = 0.1761$.

The two given values are $\log 2 = 0.3010$ and $\log 3 = 0.4771$, and $\log 10 = 1$.

(i) $\log 6 = \log(2\times3) = \log 2 + \log 3 = 0.3010 + 0.4771 = 0.7781$ (product law).

(ii) $\log 12 = \log(2^{2}\times3) = 2\log 2 + \log 3 = (2\times0.3010) + 0.4771 = 0.6020 + 0.4771 = 1.0791$ (power and product laws).

(iii) $\log 5 = \log\!\left(\dfrac{10}{2}\right) = \log 10 - \log 2 = 1 - 0.3010 = 0.6990$ (quotient law).

(iv) $\log 1.5 = \log\!\left(\dfrac{3}{2}\right) = \log 3 - \log 2 = 0.4771 - 0.3010 = 0.1761$ (quotient law).

Cross-check by an independent decomposition. (ii) $12 = 6\times2$, so $\log 12 = \log 6 + \log 2 = 0.7781 + 0.3010 = 1.0791$ — identical to the $2^{2}\times3$ route. (iii) $5 = 10/2$ can also be checked as $\log 5 = \log(3\times10/6) = \log 3 + 1 - \log 6 = 0.4771 + 1 - 0.7781 = 0.6990$ — the same value, confirming the consistency of the results.

A frequent error is writing $\log 6 = \log 2 \times \log 3$ (multiplying the logarithms) instead of $\log 2 + \log 3$ (adding them). The product law converts a product of numbers into a sum of their logarithms, never a product of logarithms.