(a) Given that \(3 \times 9^{1 + x} = 27^{-x}\), find x.
(b) Evaluate \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)
(b) Evaluate \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) \(3 \times 9^{1 + x} = 27^{-x}\)
\(3 \times (3^{2})^{1 + x} = (3^{3})^{-x}\)
\(3 \times 3^{2 + 2x} = 3^{-3x} \)
\(3^{2x + 2 + 1} = 3^{-3x} \implies 3^{2x + 3} = 3^{-3x}\)
\(\implies 2x + 3 = - 3x\)
\(-5x = 3 \implies x = \frac{-3}{5}\)
(b) \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)
= \(\log_{10} (\frac{\sqrt{35} \times \sqrt{2}}{\sqrt{7}}\)
= \(\log_{10} (\frac{\sqrt{35 \times 2}{\sqrt{7}}\)
= \(\log_{10} \sqrt{10}\)
= \(\log_{10} 10^{\frac{1}{2}}\)
= \(\frac{1}{2} \log_{10} 10 = \frac{1}{2}\)
(a) \(3 \times 9^{1 + x} = 27^{-x}\)
\(3 \times (3^{2})^{1 + x} = (3^{3})^{-x}\)
\(3 \times 3^{2 + 2x} = 3^{-3x} \)
\(3^{2x + 2 + 1} = 3^{-3x} \implies 3^{2x + 3} = 3^{-3x}\)
\(\implies 2x + 3 = - 3x\)
\(-5x = 3 \implies x = \frac{-3}{5}\)
(b) \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)
= \(\log_{10} (\frac{\sqrt{35} \times \sqrt{2}}{\sqrt{7}}\)
= \(\log_{10} (\frac{\sqrt{35 \times 2}{\sqrt{7}}\)
= \(\log_{10} \sqrt{10}\)
= \(\log_{10} 10^{\frac{1}{2}}\)
= \(\frac{1}{2} \log_{10} 10 = \frac{1}{2}\)