2015 - WAEC Mathematics Past Questions and Answers - page 1

\(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\)

\(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0\)

\(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0\)

\(3^{(3x + 1 - x - 4x)} = 3^0\)

\(3^{(1 - 2x)} = 3^0\)

since the bases are equal,

1 - 2x = 0

- 2x = -1

x = \(\frac{1}{2}\)

If log_x 64 = 3, then 64 = x³

4³ = x³

Since the indices are equal,

x = 4

Hence, x log₂8 = log₂8^x

= log₂8⁴

= log₂(2³)₄

= log₂2¹²

= 1 log₂ = 1(1)

= 12

If 2ⁿ = y,

then, 2\(^{(2 + \frac{n}{3})}\) = 2² x 2\(^\frac{n}{3}\)

= 4 x (2ⁿ)\(^{\frac{1}{3}}\)

But y = 2ⁿ, hence

2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\)

= 4y\(^\frac{1}{3}\)

6ax - 12by - 9ay + 8bx

= 6ax - 9ay + 8bx - 12by

= 3a(2x - 3y) + 4b(2x - 3y)

= (3a + 4b)(2x - 3y)

Let x = \(\frac{3}{4}\) or x = -4

i.e. 4x = 3 or x = -4

(4x - 3)(x + 4) = 0

therefore, 4x² + 13x - 12 = 0

If m = 4, n = 9, r = 16,

then \(\frac{m}{n}\) - 1\(\frac{7}{9}\) + \(\frac{n}{r}\)

= \(\frac{4}{9}\) - \(\frac{16}{9}\) + \(\frac{9}{16}\)

= \(\frac{64 - 258 + 81}{144}\)

= \(\frac{-111}{144}\)

= - 1\(\frac{37}{48}\)