Grade 12 Examination Question Logarithm


To solve for \( x \) in the equation:

\[ \log_b x = \frac{2}{3} \log_b 27 + 2 \log_b 2 - \log_b 3, \]

we can use the properties of logarithms to simplify and solve for \( x \).

Step-by-Step Solution:

  1. Apply the Power Rule:

    The power rule of logarithms states that \( \log_b a^n = n \log_b a \). Apply this to each term:

    \[ \log_b x = \frac{2}{3} \log_b 27 + 2 \log_b 2 - \log_b 3 \]

    This can be rewritten as:

    \[ \log_b x = \log_b 27^{\frac{2}{3}} + \log_b 2^2 - \log_b 3 \]

  2. Simplify the Exponents:

    Calculate \( 27^{\frac{2}{3}} \) and \( 2^2 \):

    \[ 27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{3 \times \frac{2}{3}} = 3^2 = 9 \]

    \[ 2^2 = 4 \]

    So the equation becomes:

    \[ \log_b x = \log_b 9 + \log_b 4 - \log_b 3 \]

  3. Combine the Logarithms:

    Use the product and quotient rules of logarithms:

    \[ \log_b x = \log_b (9 \times 4) - \log_b 3 = \log_b 36 - \log_b 3 \]

    \[ \log_b x = \log_b \left(\frac{36}{3}\right) = \log_b 12 \]

  4. Solve for \( x \):

    Since \( \log_b x = \log_b 12 \), it follows that:

    \[ x = 12 \]

Answer:

\[ \boxed{x = 12} \]


Also check out

Grade 12 Examination Question - Trigonometry


Previous Post Next Post

Advertisement