Surds and indices(level - 2)
Q1–10 of 50If $12^{12x} \times 4^{24x + 12} \times 5^{2y} = 8^{4z} \times 20^{12x} \times 243^{3x - 6}$ where $x, y,$ and $z$ are natural numbers, then find the value of $x + y + z$.
Break down both sides into prime factors ($2, 3, 5, 7$):
Left Side:
$12^{12x} = (2^2 \cdot 3)^{12x} = 2^{24x} \cdot 3^{12x}$
$42^{4x + 12} = (2 \cdot 3 \cdot 7)^{4x + 12} = 2^{4x+12} \cdot 3^{4x+12} \cdot 7^{4x+12}$
$5^{2y}$
Total Left: $2^{28x + 12} \cdot 3^{16x + 12} \cdot 5^{2y} \cdot 7^{4x + 12}$
Right Side:
$8^{4z} = (2^3)^{4z} = 2^{12z}$
$20^{12x} = (2^2 \cdot 5)^{12x} = 2^{24x} \cdot 5^{12x}$
$243^{3x - 6} = (3^5)^{3x - 6} = 3^{15x - 30}$
Total Right: $2^{24x + 12z} \cdot 3^{15x - 30} \cdot 5^{12x}$
Find the sum of all real values of $k$ for which $\left(\frac{1}{8}\right)^k \times \left(\frac{1}{32768}\right)^{\frac{1}{3}} = \frac{1}{8} \times \left(\frac{1}{32768}\right)^{\frac{1}{k}}$.
Equate the exponents:
Multiply through by $k$ to form a quadratic equation:
Sum of roots for $ax^2 + bx + c = 0$ is $-\frac{b}{a}$:
Let $a, b, m$ and $n$ be natural numbers such that $a > 1$ and $b > 1$. If $a^m b^n = 144^{145}$, then find the largest possible value of $(n - m)$.
To maximize $(n - m)$, make $n$ as large as possible and $m$ as small as possible:
Maximize $n$: Choose the smallest base $b = 2$, so $b^n = 2^{580} \implies \mathbf{n = 580}$.
Minimize $m$: The leftover part is $a^m = 3^{290}$. Write this as $(3^{290})^1$, so $a = 3^{290}$ and $\mathbf{m = 1}$.
Subtract $m$ from $n$:
If $x = 4096^{7+4\sqrt{3}}$, then which of the following expressions evaluates exactly to $64$?
We want $x^k = 64^1$, which means $k \cdot (14 + 8\sqrt{3}) = 1$.
Rationalize the fraction:
Therefore:
If $a, b, c$ are non-zero real numbers and $14^a = 36^b = 84^c$, then find the value of $\frac{6b}{c} - \frac{6b}{a}$.
Look for a relationship between the numbers $14, 36/6, 84$:
Substitute the expressions in terms of $k$:
Equate the exponents:
Multiply both sides by $6b$:
If $5.55^x = 0.555^y = 1000$, then find the value of $\frac{1}{x} - \frac{1}{y}$.
Divide the first equation by the second:
Equate exponents:
Given that $x^{2018}y^{2017} = \frac{1}{2}$ and $x^{2016}y^{2019} = 8$, find the value of $x^2 + y^3$.
Multiply the two original equations together:
Solving for individual values gives $x^2 = \frac{1}{4}$ and $y = 2 \implies y^3 = 8$.
Calculate $x^2 + y^3$:
Given that $x^{2018}y^{2017} = \frac{1}{2}$ and $x^{2016}y^{2019} = 8$, find the value of $x^2 + y^3$.
Multiply the two original equations together:
Solving for individual values gives $x^2 = \frac{1}{4}$ and $y = 2 \implies y^3 = 8$.
Calculate $x^2 + y^3$:
Find the value of the expression $\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \dots + \sqrt{1 + \frac{1}{2007^2} + \frac{1}{2008^2}}$.
Apply to the sum:
Telescoping cancellation:
If $x = \sqrt{7 + \sqrt{7 + \sqrt{7 + \dots}}}$, then which of the following expressions defines the boundary layout of $x$?
Set $x = \sqrt{7 + x}$.
Square both sides: $x^2 - x - 7 = 0$.
Use the quadratic formula:
$$x = \frac{1 + \sqrt{1 + 28}}{2} = \frac{1 + \sqrt{29}}{2}$$Since $5 < \sqrt{29} < 6$:
$$\frac{1 + 5}{2} < x < \frac{1 + 6}{2} \implies 3 < x < 3.5$$
Correct Answer: (B) 3 < x < 4