(a) Given that \(3 \times 9^{1 + x} = 27^{-x}\), find x. (b) Evaluate \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)

Assessment: WAEC SSCE - General Mathematics - 1994 (Essay) Subject: General Mathematics

Question 1 Report

(a) Given that \(3 \times 9^{1 + x} = 27^{-x}\), find x.

(b) Evaluate \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)

Answer Details

(a) Express everything as powers of 3: \(9 = 3^2\), \(27 = 3^3\).

\(3\times 9^{1+x} = 3^1 \times 3^{2(1+x)} = 3^{1 + 2 + 2x} = 3^{3 + 2x}\)

\(27^{-x} = 3^{-3x}\)

Equating indices: \(3 + 2x = -3x \ \Rightarrow\ 5x = -3 \ \Rightarrow\ \mathbf{x = -\dfrac{3}{5}}\).

(b) Use \(\log\sqrt{a} = \tfrac{1}{2}\log a\) and combine:

\(\log_{10}\sqrt{35} + \log_{10}\sqrt{2} - \log_{10}\sqrt{7} = \log_{10}\!\left(\dfrac{\sqrt{35}\times\sqrt{2}}{\sqrt{7}}\right)\)

\(= \log_{10}\sqrt{\dfrac{35\times 2}{7}} = \log_{10}\sqrt{10} = \log_{10} 10^{1/2} = \dfrac{1}{2}\)

Value \(= \mathbf{\dfrac{1}{2}}\).

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