Exponential equations

Fluency

Initial value $a = 7$ , base $b = 2$ .
$x = 0$ : $y = 1$ . $x = 1$ : $y = 4$ . $x = 3$ : $y = 64$ .
Decay (since $0 < 0.6 < 1$ ).
$y = 100 \times 1.2^t$ .
$81 = 3^4$ , so $x = 4$ .
$\tfrac{1}{8} = 2^{-3}$ , so $x = -3$ .
$10000 = 10^4$ , so $x = 4$ .
Number of half-lives: $\tfrac{12}{4} = 3$ . Mass: $160 \times \left(\tfrac{1}{2}\right)^3 = 160 \times \tfrac{1}{8} = 20$ g.

Reasoning

$64 = 4^3$ , so $x = 3$ .
$25 = 5^2$ , $125 = 5^3$ . $(5^2)^x = 5^3$ . $2x = 3$ . $x = \tfrac{3}{2}$ .
Model: $N = 800 \times 3^{t/5}$ . After $15$ days: $N = 800 \times 3^3 = 800 \times 27 = 21600$ insects.
Half-lives: $\tfrac{18}{6} = 3$ . Mass: $500 \times \left(\tfrac{1}{2}\right)^3 = 500 \times \tfrac{1}{8} = 62.5$ g. Model: $M = 500 \times \left(\tfrac{1}{2}\right)^{t/6}$ .
Rule of 70: $\tfrac{70}{8} = 8.75$ years. Check: $1000 \times 1.08^9 = 1000 \times 1.999 \approx 1999$ dollars. Confirms doubling in about $9$ years.
$5^x = 125 = 5^3$ . $x = 3$ .

Reasoning

Since $b > 0$ , raising $b$ to any power gives a positive result: $b^x > 0$ for all $x$ . Multiplying by $a > 0$ keeps it positive. So $y = a \times b^x > 0$ always. The curve approaches the $x$ -axis (its asymptote) but never touches or crosses it.
Colony A: $N_A = 100 \times 2^{t/2}$ . Colony B: $N_B = 3200 \times \left(\tfrac{1}{2}\right)^{t/2}$ . Set equal: $100 \times 2^{t/2} = 3200 \times 2^{-t/2}$ . Multiply both sides by $2^{t/2}$ : $100 \times 2^t = 3200$ . $2^t = 32 = 2^5$ . $t = 5$ . They are equal after $\mathbf{10}$ hours. Wait — let $u = t/2$ : $100 \times 2^u = 3200 \times 2^{-u}$ , $100 \times 2^{2u} = 3200$ , $2^{2u} = 32 = 2^5$ , $2u = 5$ , $u = 2.5$ , $t = 5$ hours. Check: $N_A = 100 \times 2^{2.5} \approx 566$ , $N_B = 3200 \times 2^{-2.5} \approx 566$ . Equal after 5 hours.
$V = 30000 \times 0.88^t$ . Need $V < 10000$ : $0.88^t < \tfrac{1}{3}$ . By trial: $0.88^8 \approx 0.3596$ , $0.88^9 \approx 0.3165$ , $0.88^{10} \approx 0.2785$ . After $9$ years: $V \approx 9494 < 10000$ . First worth less than $10000 after 9 whole years.
Half-life $= 5$ years (the denominator in the exponent). After $20$ years: $M = 800 \times \left(\tfrac{1}{2}\right)^4 = 800 \times \tfrac{1}{16} = 50$ g. Below $100$ g: $\left(\tfrac{1}{2}\right)^{t/5} < \tfrac{1}{8} = \left(\tfrac{1}{2}\right)^3$ , so $\tfrac{t}{5} > 3$ , $t > 15$ years.

Reasoning

$8 = 2^3$ and $4 = 2^2$ . So $2^{3(2x+1)} = 2^{2(3x-2)}$ . $6x + 3 = 6x - 4$ . $3 = -4$ , which is false. No solution.
Let $P_n$ be the population after year $n$ . $P_0 = 10000$ . $P_{n+1} = 0.9 P_n + 500$ . Year 1: $0.9(10000) + 500 = 9500$ . Year 2: $0.9(9500) + 500 = 9050$ . Year 3: $0.9(9050) + 500 = 8645$ . The population is declining. (It would stabilise at $P = 0.9P + 500$ , so $0.1P = 500$ , $P = 5000$ , but it has not reached that level yet.)
Need $5000 \times 1.06^t > 8000 \times 1.03^t$ . $\left(\tfrac{1.06}{1.03}\right)^t > 1.6$ . $1.02913^t > 1.6$ . By trial: $t = 16$ : $1.02913^{16} \approx 1.585$ ; $t = 17$ : $\approx 1.631$ . Investment A first exceeds B after 17 whole years.

Year 10 core - answers