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Question

Find the value of $$\log 6 + 2 \log 5 + \log 4 – \log 3 – \log 2$$.


Solution

$$\log 6 + 2 \log 5 + \log 4 – \log 3 – \log 2$$
$$= (\log 6 + \log 5^2 + \log 4) – (\log 3 + \log 2)$$
$$= \log (6 \times 5^{2} \times 4) – \log 3 \times 2$$
$$=\log\dfrac{6\times5^{2}\times4}{3*2}$$
$$= \log (5^{2} \times 4)$$
$$= \log (25 \times 4) = \log 100 = 2$$
Hence, $$\log 6 + 2 \log 5 + \log 4 – \log 3 – \log 2 = 2$$

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