Indices, Surds & Identities: CAT previous year questions

Solution

Left side $= 3^{1^2 + 2^2 + \dots + 20^2} = 3^{2870}$.

Right side $= 3^{21 + 22 + \dots + (20 + m)} = 3^{20m + \frac{m(m + 1)}{2}}$.

Need $20m + \dfrac{m(m + 1)}{2} > 2870$. $m = 57$: $1140 + 1653 = 2793$ ✗. $m = 58$: $1160 + 1711 = 2871$ ✓.

Smallest $m = 58$.

Solution

Let $u = 3^{x^2 + 2x - 3}$. Then $9^{x^2 + 2x - 3} = u^2$ and $3^{x^2 + 2x - 2} = 3u$.

$u^2 - 12u + 27 = 0 \Rightarrow u = 3$ or $9$.

$x^2 + 2x - 3 = 1 \Rightarrow x^2 + 2x - 4 = 0$ (product of roots $-4$).

$x^2 + 2x - 3 = 2 \Rightarrow x^2 + 2x - 5 = 0$ (product of roots $-5$).

Product of all four values $= (-4)(-5) = 20$.

Solution

$\left(x + \dfrac{1}{x}\right)^2 = 25 + 2 = 27 \Rightarrow x + \dfrac{1}{x} = 3\sqrt{3}$ (as $x > 0$).

$x^3 + \dfrac{1}{x^3} = (3\sqrt{3})^3 - 3(3\sqrt{3}) = 81\sqrt{3} - 9\sqrt{3} = 72\sqrt{3}$.

$x^4 + \dfrac{1}{x^4} = 25^2 - 2 = 623$.

$x^7 + \dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right)\left(x^4 + \dfrac{1}{x^4}\right) - \left(x + \dfrac{1}{x}\right) = 72\sqrt{3} \times 623 - 3\sqrt{3} = 44853\sqrt{3}$.

Solution

Write each side in primes.

Left: $2^{24x} 3^{12x} \cdot 2^{48x + 24} \cdot 5^{2y} = 2^{72x + 24}\, 3^{12x}\, 5^{2y}$.

Right: $2^{12z} \cdot 2^{24x} 5^{12x} \cdot 3^{15x - 30}$.

Powers of 3: $12x = 15x - 30 \Rightarrow x = 10$. Powers of 5: $2y = 120 \Rightarrow y = 60$.

Powers of 2: $744 = 12z + 240 \Rightarrow z = 42$.

$x + y + z = 112$.

Solution

$32768 = 8^5$. Write everything as powers of 8:

$8^{-k} \cdot 8^{-5/3} = 8^{-1} \cdot 8^{-5/k} \Rightarrow k + \dfrac{5}{3} = 1 + \dfrac{5}{k}$.

$3k^2 + 2k - 15 = 0$ (two real roots).

Sum of values $= -\dfrac{2}{3}$.

Solution

$29 - 12\sqrt{5} = (2\sqrt{5} - 3)^2$, so the positive root is $-3 + 2\sqrt{5}$.

$2\sqrt{5}$ can be written as $b\sqrt{n}$ with $b = 2, n = 5$ or $b = 1, n = 20$.

$a + b + n$ is $4$ or $18$. Maximum $= 18$.

Solution

Let $A = x + 6\sqrt{2}$, $B = x - 6\sqrt{2}$, so $A - B = 12\sqrt{2}$.

$\sqrt{A} - \sqrt{B} = 2\sqrt{2} \Rightarrow \sqrt{A} + \sqrt{B} = \dfrac{12\sqrt{2}}{2\sqrt{2}} = 6$.

$\sqrt{A} = 3 + \sqrt{2} \Rightarrow A = 11 + 6\sqrt{2}$.

$x = 11$.

Solution

$(a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3}$.

$2ab = 30 \Rightarrow ab = 15$, and $a^2 + 3b^2 = 52$.

$a = 5, b = 3$: $25 + 27 = 52$ ✓.

$a + b = 8$.

Solution

$(5x + 9) - (5x - 9) = 18$, so $\sqrt{5x + 9} - \sqrt{5x - 9} = \dfrac{18}{3(2 + \sqrt{2})} = 3(2 - \sqrt{2})$.

Adding to the given equation: $2\sqrt{5x + 9} = 12 \Rightarrow 5x + 9 = 36 \Rightarrow 5x = 27$.

$\sqrt{10x + 9} = \sqrt{63} = 3\sqrt{7}$.

Solution

Let $u = 2^{2x^2}$ and $v = 2^{x + 15}$. Then $2^{4x^2} = u^2$, $2^{2x^2 + x + 16} = 2uv$, $2^{2x + 30} = v^2$.

$u^2 - 2uv + v^2 = 0 \Rightarrow u = v \Rightarrow 2x^2 = x + 15$.

$2x^2 - x - 15 = 0$: sum of roots $= \dfrac{1}{2}$.

Solution

$x^8 + x^{-8} = 47 \Rightarrow x^4 + x^{-4} = 7 \Rightarrow x^2 + x^{-2} = 3 \Rightarrow x + x^{-1} = \sqrt{5}$.

$x^3 + x^{-3} = 5\sqrt{5} - 3\sqrt{5} = 2\sqrt{5}$; $x^6 + x^{-6} = 20 - 2 = 18$.

$x^9 + x^{-9} = (x^3 + x^{-3})(x^6 - 1 + x^{-6}) = 2\sqrt{5} \times 17 = 34\sqrt{5}$.

Solution

$\left(\dfrac{4a}{b}\right)^{6x - y} = \left(\dfrac{2a}{b}\right)^{y - 6x}$ for all $a, b$ forces $6x - y = 0$, so $y = 6x$.

$\dfrac{875}{2401} = \dfrac{5^3 \cdot 7}{7^4} = \left(\dfrac{7}{5}\right)^{-3}$, and $\left(\sqrt{\dfrac{7}{5}}\right)^{3x - y} = \left(\dfrac{7}{5}\right)^{\frac{3x - y}{2}}$.

$\dfrac{3x - y}{2} = -3 \Rightarrow 3x - 6x = -6 \Rightarrow x = 2$, $y = 12$.

$x + y = 14$.

Solution

Product $= 10^{\frac{1 + 2 + \dots + n}{7}} = 10^{\frac{n(n + 1)}{14}}$.

We need this to exceed 999. $n = 5$: $10^{15/14} \approx 11.8$ ✗. $n = 6$: $10^{3} = 1000$ ✓.

Smallest $n = 6$.

Solution

$4096 = 64^2$, so $x = 64^{2(7 + 4\sqrt{3})}$. We need $x^p = 64$, i.e. $p = \dfrac{1}{2(7 + 4\sqrt{3})}$.

Since $(7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 1$: $p = \dfrac{7 - 4\sqrt{3}}{2} = \dfrac{7}{2} - 2\sqrt{3}$.

So $64 = \dfrac{x^{7/2}}{x^{2\sqrt{3}}}$.