\( \begin{array}{ll}\text { (c) }-2^{x}=-16 & \text { (t) }-2^{x}=-32 \\ \text { (ii) }-2^{x}=-1 & \text { (i) } 7^{x}=49 \\ \text { (k) } 7^{x}=1 & \text { (l) } 3^{2 x}=81 \\ \text { (ii) } 10^{10 x}=100 & \text { (o) } 4^{x}=2 \\ \text { (4) } 25^{x}=125 & \text { (r) } 9^{3 x}=27^{x} \\ \text { (t) } 9^{3 x-1}=27 & \text { (u) } 16^{2 x}=4^{x-1}\end{array} \)

Below are one acceptable set of solutions. In each case we assume that, for example,   –2ˣ means –(2ˣ) rather than (–2)ˣ. • (c) –2ˣ = –16   Add a minus sign to both sides: 2ˣ = 16   Since 16 = 2⁴, we have 2ˣ = 2⁴ so x = 4. • (t) –2ˣ = –32   Then 2ˣ = 32, and 32 = 2⁵ so x = 5. • (ii) –2ˣ = –1   Then 2ˣ = 1. Since 2⁰ = 1, we have x = 0. • (i) 7ˣ = 49   Write 49 as 7² so that 7ˣ = 7²; hence x = 2. • (k) 7ˣ = 1   Since 7⁰ = 1, it follows that x = 0. • (l) 3^(2x) = 81   Notice that 81 = 3⁴. Therefore 3^(2x) = 3⁴ implying 2x = 4 and so x = 2. • (ii) 10^(10x) = 100   Write 100 as 10². Then 10^(10x) = 10² so 10x = 2 and x = 0.2. • (o) 4ˣ = 2   Write 4 as 2² so that 4ˣ = (2²)ˣ = 2^(2x). Hence the equation becomes 2^(2x) = 2¹, so 2x = 1 and x = ½. • (4) 25ˣ = 125   Express 25 = 5² and 125 = 5³. The equation becomes (5²)ˣ = 5³, i.e. 5^(2x) = 5³. Thus, 2x = 3 and x = 1.5 (or 3/2). • (r) 9^(3x) = 27^(x)   Write 9 = 3² and 27 = 3³. Then    9^(3x) = (3²)^(3x) = 3^(6x)    27^(x) = (3³)^(x) = 3^(3x).   So 3^(6x) = 3^(3x) which implies 6x = 3x, hence 3x = 0 and x = 0. • (t) 9^(3x–1) = 27   Again, 9 = 3² and 27 = 3³. Rewrite the left side:    9^(3x–1) = (3²)^(3x–1) = 3^(2(3x–1)) = 3^(6x–2).   Thus 3^(6x–2) = 3³, so 6x – 2 = 3, giving 6x = 5 and x = 5/6. • (u) 16^(2x) = 4^(x–1)   Write 16 as 4² so that:    16^(2x) = (4²)^(2x) = 4^(4x).   Then the equation becomes 4^(4x) = 4^(x–1). Equate exponents: 4x = x – 1, so    4x – x = –1 → 3x = –1, hence x = –1/3. Thus, the answers are:   (c) x = 4   (t) x = 5   (ii) x = 0    [for –2ˣ = –1]   (i) x = 2   (k) x = 0   (l) x = 2   (ii) x = 0.2  [for 10^(10x) = 100]   (o) x = 1/2   (4) x = 3/2   (r) x = 0   (t) x = 5/6   (u) x = –1/3 Any answer set equivalent to these is correct.