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Given that 3 = 0.4771 and 2 = 0.3010. We assume the logarithm is base 10.

Chemistry

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Answer

-0.1428

Step-by-step solution

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Given that log⁡3=0.4771\log 3 = 0.4771 and log⁡2=0.3010\log 2 = 0.3010. We assume the logarithm is base 10.

Problem (i): Calculate log⁡0.72\log 0.72.

  1. Express 0.72 as a fraction and then in terms of prime factors.

    0.72=72100=23×321020.72 = \frac{72}{100} = \frac{2^3 \times 3^2}{10^2}

  2. Apply logarithm properties log⁡ab=log⁡a−log⁡b\log \frac{a}{b} = \log a - \log b and log⁡(xy)=log⁡x+log⁡y\log (xy) = \log x + \log y.

    log⁡0.72=log⁡(23×32102)=log⁡(23×32)−log⁡(102)\log 0.72 = \log \left(\frac{2^3 \times 3^2}{10^2}\right) = \log (2^3 \times 3^2) - \log (10^2) =log⁡(23)+log⁡(32)−log⁡(102)= \log (2^3) + \log (3^2) - \log (10^2)

  3. Apply the logarithm property log⁡ab=blog⁡a\log a^b = b \log a.

    =3log⁡2+2log⁡3−2log⁡10= 3 \log 2 + 2 \log 3 - 2 \log 10

  4. Substitute the given values and log⁡10=1\log 10 = 1 (since we assume base 10).

    =3(0.3010)+2(0.4771)−2(1)= 3(0.3010) + 2(0.4771) - 2(1) =0.9030+0.9542−2= 0.9030 + 0.9542 - 2 =1.8572−2= 1.8572 - 2 =−0.1428= -0.1428

    The value of log⁡0.72\log 0.72 is −0.1428\boxed{-0.1428}.

    Problem (ii): Calculate log⁡56\log \sqrt{56}.

  5. Rewrite the square root as an exponent

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