Solve 1a and 1b

Mathematics

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Answer

log⁡10(103)\log_{10} \left(\frac{10}{3}\right)

−log⁡1015log⁡10(95)\frac{- \log_{10} 15}{\log_{10} \left(\frac{9}{5}\right)}

Step-by-step solution

9 steps

  1. Apply the power rule of logarithms, alog⁡bx=log⁡bxaa \log_b x = \log_b x^a.

    log⁡103612−log⁡1032+log⁡105\log_{10} 36^{\frac{1}{2}} - \log_{10} 3^2 + \log_{10} 5 log⁡1036−log⁡109+log⁡105\log_{10} \sqrt{36} - \log_{10} 9 + \log_{10} 5 log⁡106−log⁡109+log⁡105\log_{10} 6 - \log_{10} 9 + \log_{10} 5

  2. Apply the quotient rule of logarithms, log⁡bx−log⁡by=log⁡b(xy)\log_b x - \log_b y = \log_b \left(\frac{x}{y}\right).

    log⁡10(69)+log⁡105\log_{10} \left(\frac{6}{9}\right) + \log_{10} 5 log⁡10(23)+log⁡105\log_{10} \left(\frac{2}{3}\right) + \log_{10} 5

  3. Apply the product rule of logarithms, log⁡bx+log⁡by=log⁡b(xy)\log_b x + \log_b y = \log_b (xy).

    log⁡10(23×5)\log_{10} \left(\frac{2}{3} \times 5\right) log⁡10(103)\log_{10} \left(\frac{10}{3}\right)

    The simplified expression is:

    log⁡10(103)\boxed{\log_{10} \left(\frac{10}{3}\right)}

    b)

  4. Take the logarithm of both sides.

    We will use log⁡10\log_{10}.

    log⁡10(32x+1)=log⁡10(5x−1)\log_{10} (3^{2x+1}) = \log_{10} (5^{x-1})

  5. Apply the power rule of logarithms, log⁡bxa=alog⁡bx\log_b x^a = a \log_b x.

    (2x+1)log⁡103=(x−1)log⁡105(2x+1) \log_{10} 3 = (x-1) \log_{10} 5

  6. Expand both sides of the equation.

    2xlog⁡103+log⁡103=xlog⁡105−log⁡1052x \log_{10} 3 + \log_{10} 3 = x \log_{10} 5 - \log_{10} 5

  7. Group terms containing xx on one side and constant terms on the other side.

    2xlog⁡103−xlog⁡105=−log⁡105−log⁡1032x \log_{10} 3 - x \log_{10} 5 = - \log_{10} 5 - \log_{10} 3

  8. Factor out xx from the terms on the left side.

    x(2log⁡103−log⁡105)=−(log⁡105+log⁡103)x (2 \log_{10} 3 - \log_{10} 5) = - (\log_{10} 5 + \log_{10} 3)

  9. Solve for xx.

    x=−(log⁡105+log⁡103)2log⁡103−log⁡105x = \frac{- (\log_{10} 5 + \log_{10} 3)}{2 \log_{10} 3 - \log_{10} 5}

    Using logarithm properties, this can also be written as:

    x=−log⁡10(5×3)log⁡10(32)−log⁡105x = \frac{- \log_{10} (5 \times 3)}{\log_{10} (3^2) - \log_{10} 5} x=−log⁡1015log⁡109−log⁡105x = \frac{- \log_{10} 15}{\log_{10} 9 - \log_{10} 5} x=−log⁡1015log⁡10(95)x = \frac{- \log_{10} 15}{\log_{10} \left(\frac{9}{5}\right)}

    The value of xx is:

    −log⁡1015log⁡10(95)\boxed{\frac{- \log_{10} 15}{\log_{10} \left(\frac{9}{5}\right)}}
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