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Log\(_2\)x + 3Log\(_2\)x = 16 Add, since the Logs are like terms = 4Log\(_2\)x = 16 = Log\(_2\)x = 4 = x = 2\(^4\) = 16.

To rationalise \(\frac{\sqrt{2}}{3 + \sqrt{2}}\), multiply both the numerator and denominator of the fraction by the conjugate of the denominator, that is, 3 - \(\sqrt{2}\) \(\frac{\sqrt{2}}{3 + \sqrt{2}}\) x \(\frac{3 - \sqrt{2}}{3 - \sqrt{2}}\) = \(\frac{3\sqrt{2} - 2}{9 - 2}\) = \(\frac{3\sqrt{2} - 2}{7}\)

If Log\(_k243\) = 5 Convert to index form. k\(^5\) = 243 k\(^5\) = 3\(^5\) Since the powers on both sides are equal, equate the bases. k = 3

\(\left(\frac{1}{2}\right)^{t - 1} = 64,\) rewrite \(64\) as a power of \(\frac{1}{2}\): \(64 = \left(\frac{1}{2}\right)^{-6}.\) Equate the exponents: \(t - 1 = -6\) Thus, \(t = -5\)

0.00001 = 10\(^{-5}\), so, log\(_{10}\)(10\(^{-5}\)) = - 5. Then -(-5) = 5.

log\(_{10}\)x = 2.3675 ⇒ x = 10\(^{2.3675}\) ⇒ x ≈ 0.02331 log\(_{10}\)y = 2.9738 ⇒ y = 10\(^{2.9738}\) ⇒ y ≈ 0.09414 ∴ x + y = 0.02331 + 0.09414 = 0.11745 = 0.118 (3 s.f.)

\(\sqrt{75} - \sqrt{12} + \sqrt{27} = y\sqrt{3}\) Simplify each term: - \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\) - \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) - \(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\) Substitute: \(5\sqrt{3} - 2\sqrt{3} + 3\sqrt{3} = y\sqrt{3}\) Combine like terms: \((5 - 2 + 3)\sqrt{3} = y\sqrt{3}\) \(6\sqrt{3} = y\sqrt{3}\) Thus, y = 6

\(\sqrt[4]{0.16} = (0.16)^{\frac{1}{4}} = (16 \times 10^{-2})^{\frac{1}{4}} = 2 \times 10^{-\frac{1}{2}}\)

Given: 125\(^{−\frac{1}{3}}\) \times 49\(^{−\frac{1}{2}}\) \times 10\(^0\) 125\(^{-1/3}\) = (5\(^3\))\(^{-1/3}\) = 5\(^{-1}\) = \(\frac{1}{5}\). 49\(^{-1/2}\) = (7\(^2\))\(^{-1/2}\) = 7\(^{-1}\) = \(\frac{1}{7}\). 10\(^0\) = 1. Product = \(\frac{1}{5}\) x \(\frac{1}{7}\) x 1 = \(\frac{1}{35}\).

0.000834 = 8.34 x 10\(^{-4}\) Move the decimal point 4 places to your right.

0.027\(^{-\frac{1}{3}}\) = (\(\frac{27}{100}\))\(^{-\frac{1}{3}}\) = (\(\frac{10^3}{3^3}\))\(^{\frac{1}{3}}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)

log\(_2a\) = log\(_84\) Let log\(_84\) = y Then 4 = 8\(^y\) 2\(^2\) = 2\(^{3y}\) 2 = 3y y = \(\frac{2}{3}\) log\(_2a\) = log\(_84\) = \(\frac{2}{3}\) log\(_2a\) = \(\frac{2}{3}\) a = 2\(^{\frac{2}{3}}\) (Note: The logarithm was converted to index form.)

ai From the four-figure/ mathematical table 2sin63.35º = 2(0.8938) = 1.7876 ii. From the four-figure/ mathematical table, log cos 44.74º = log 0.7103 = \(\bar{1}\).8515 iii. log k - log(k - 2) = log 5 = log \(\frac{\text{k}}{k - 2}\) = log 5 (law of logarithms) log \(\frac{\text{k}}{k - 2}\) = log 5 \(\frac{\text{k}}{k - 2}\) = 5 cross multiply 5(k - 2) = k 5k - 10 = k 5k - k = 10 4k = 10 k = \(\frac{10}{4}\) = \(\frac{5}{2}\). b. \(\frac{3.68^2 \times 6.705}{\sqrt{0.3581}}\) = 151.8. See the steps in the diagram above.

\(\frac{16^{2 + x}}{4} = 64^x\) 16\(^{2 + x} = 4 \times 64^x\) 4\(^{2(2 + x)} = 4^1 \times 4^{3(x)}\) (Equate coefficients since the bases are the same) 2(2 + x) = 1 + 3x 4 + 2x = 1 + 3x 2x - 3x = 1 - 4 -x = -3 x = 3.

Length of Arc = \(\frac{\theta}{360}\) x 2\(\pi\)r Length of arc = 33cm, diameter = 28cm → r = 14cm 33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7}\times14\) \(\theta\) = \(\frac{33 \times 180}{22 \times 2}\) \(\theta\) = 135º