Question paper, Set 9

$(-1)^{37} = -1$ (37 is odd), $(-1)^{52} = 1$ (52 is even), $7^{0} = 1$

$-1 + 1 + 1 = 1$

Answer: $1$

$\displaystyle \left(\frac{3}{5}\right)^{4} = \left(\frac{5}{3}\right)^{-4}$

$\displaystyle \left(\frac{5}{3}\right)^{-4} \times \left(\frac{5}{3}\right)^{6} = \left(\frac{5}{3}\right)^{2}$

$\displaystyle = \frac{25}{9}$

Answer: $\displaystyle \frac{25}{9}$

$= (5^{2})^{4} \times (5^{3})^{-2}$

$= 5^{8} \times 5^{-6}$

$= 5^{8 - 6} = 5^{2}$

$= 25$

Answer: $25$

$\displaystyle = \frac{(5^{2})^{\frac{3}{2}} \times (2^{3})^{\frac{5}{3}}}{(10^{4})^{\frac{3}{4}}}$

$\displaystyle = \frac{5^{3} \times 2^{5}}{10^{3}}$

$\displaystyle = \frac{125 \times 32}{1000}$

$= 4$

Answer: $4$

$\displaystyle = \left[\left(\frac{1}{3}\right)^{3}\right]^{-\frac{2}{3}} + 2^{4} - 6 \times (3^{2})^{\frac{1}{2}} \times 1$

$= \left(3\right)^{2} + 16 - 6 \times 3$

$= 9 + 16 - 18$

$= 7$

Answer: $7$

$\displaystyle = \frac{15^{\frac{2}{5}} \times 5^{\frac{1}{5}} \times 25^{\frac{1}{5}}}{10^{\frac{1}{3}} \times 6^{\frac{1}{3}}} \times \frac{3^{\frac{3}{5}} \times 15^{\frac{1}{3}}}{2^{\frac{1}{3}} \times 4^{-\frac{1}{2}}}$

$\displaystyle = \frac{(3 \times 5)^{\frac{2}{5}} \times 5^{\frac{1}{5}} \times (5^{2})^{\frac{1}{5}}}{(2 \times 5)^{\frac{1}{3}} \times (2 \times 3)^{\frac{1}{3}}} \times \frac{3^{\frac{3}{5}} \times (3 \times 5)^{\frac{1}{3}}}{2^{\frac{1}{3}} \times (2^{2})^{-\frac{1}{2}}}$

$\displaystyle = \frac{3^{\frac{2}{5}} \times 5^{\frac{2}{5}} \times 5^{\frac{1}{5}} \times 5^{\frac{2}{5}}}{2^{\frac{1}{3}} \times 5^{\frac{1}{3}} \times 2^{\frac{1}{3}} \times 3^{\frac{1}{3}}} \times \frac{3^{\frac{3}{5}} \times 3^{\frac{1}{3}} \times 5^{\frac{1}{3}}}{2^{\frac{1}{3}} \times 2^{-1}}$

$\displaystyle = 2^{-\frac{1}{3} - \frac{1}{3} - \frac{1}{3} + 1} \times 3^{\frac{2}{5} - \frac{1}{3} + \frac{3}{5} + \frac{1}{3}} \times 5^{\frac{2}{5} + \frac{1}{5} + \frac{2}{5} - \frac{1}{3} + \frac{1}{3}}$

$= 2^{0} \times 3^{1} \times 5^{1}$

$= 15$

Answer: $15$

LHS $= x^{(a-b)(a+b)} \div x^{(a-c)(a+c)} \times x^{(b-c)(b+c)}$

$= x^{a^{2}-b^{2}} \div x^{a^{2}-c^{2}} \times x^{b^{2}-c^{2}}$

$= x^{a^{2}-b^{2}-(a^{2}-c^{2})+b^{2}-c^{2}}$

$= x^{a^{2}-b^{2}-a^{2}+c^{2}+b^{2}-c^{2}} = x^{0} = 1 =$ RHS

Answer: Shown: LHS $= 1 =$ RHS

Given $a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3abc$

LHS $= x^{a^{3}+b^{3}+c^{3}} = x^{3abc}$

RHS $= \left(x^{abc}\right)^{3} = x^{3abc}$

So LHS = RHS

Answer: Shown: both sides $= x^{3abc}$

$(3^{2})^{x} = 27$

$\Rightarrow 3^{2x} = 3^{3}$

$\Rightarrow 2x = 3$

$\displaystyle \Rightarrow x = \frac{3}{2}$

Answer: $\displaystyle x = \frac{3}{2}$

$22 = 2^{1} \times 11^{1}$

$\Rightarrow 2^{3x - 5} \times 11^{x - 1} = 2^{1} \times 11^{1}$

$\Rightarrow 3x - 5 = 1$ and $x - 1 = 1$

$\Rightarrow x = 2$ and $x = 2$

$\therefore x = 2$

Answer: $x = 2$

$6^{x+1} = 6^{x} \times 6^{1}$

$= 5 \times 6 = 30$

Answer: $30$

$\displaystyle \left(\frac{5}{3}\right)^{x} = \left(\frac{3}{5}\right)^{-x}$

$\displaystyle \Rightarrow \left(\frac{3}{5}\right)^{-x} \cdot \left(\frac{3}{5}\right)^{3x} = \left(\frac{3}{5}\right)^{4}$

$\displaystyle \Rightarrow \left(\frac{3}{5}\right)^{2x} = \left(\frac{3}{5}\right)^{4}$

$\Rightarrow 2x = 4 \Rightarrow x = 2$

Answer: $x = 2$

$\displaystyle 5^{x} - \frac{5^{x}}{25} = 600$

$\displaystyle \Rightarrow 5^{x}\left(1 - \frac{1}{25}\right) = 600$

$\displaystyle \Rightarrow 5^{x} \times \frac{24}{25} = 600$

$\displaystyle \Rightarrow 5^{x} = 600 \times \frac{25}{24} = 625 = 5^{4}$

$\Rightarrow x = 4$

Answer: $x = 4$

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

$\Rightarrow 3^{4x-4} = 3^{3x+3}$

$\Rightarrow 4x - 4 = 3x + 3 \Rightarrow x = 7$

$7^{x-5} = 7^{7-5} = 7^{2} = 49$

Answer: $49$

$64 = 9^{y} = (2^{x})^{y} = 2^{xy}$

$2^{xy} = 64 = 2^{6}$

$\Rightarrow xy = 6$

Answer: $xy = 6$

Let $2^{x} = 3^{y} = 24^{z} = k$ (say)

$\displaystyle \Rightarrow 2 = k^{\frac{1}{x}}$, $\displaystyle 3 = k^{\frac{1}{y}}$, $\displaystyle 24 = k^{\frac{1}{z}}$

$24 = 2^{3} \times 3$

$\displaystyle \Rightarrow k^{\frac{1}{z}} = \left(k^{\frac{1}{x}}\right)^{3} \times k^{\frac{1}{y}}$

$\displaystyle \Rightarrow k^{\frac{1}{z}} = k^{\frac{3}{x} + \frac{1}{y}}$

$\displaystyle \Rightarrow \frac{1}{z} = \frac{3}{x} + \frac{1}{y}$

$\displaystyle \Rightarrow \frac{1}{z} = \frac{3y + x}{xy}$

$\displaystyle \Rightarrow z = \frac{xy}{x + 3y}$

Answer: Shown: $\displaystyle z = \frac{xy}{x + 3y}$

$\displaystyle 0.000125 = \frac{125}{1000000} = \left(\frac{5}{100}\right)^{3} = (0.05)^{3}$

$\displaystyle (0.000125)^{\frac{2}{3}} = \left[(0.05)^{3}\right]^{\frac{2}{3}} = (0.05)^{2} = 0.0025$

Answer: $0.0025$

$(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$

$\displaystyle x^{3} = \left(2^{\frac{2}{3}}\right)^{3} + \left(2^{-\frac{2}{3}}\right)^{3} + 3 \cdot 2^{\frac{2}{3}} \cdot 2^{-\frac{2}{3}} \left(2^{\frac{2}{3}} + 2^{-\frac{2}{3}}\right)$

$\displaystyle = 4 + \frac{1}{4} + 3 \times 1 \times x$

$\displaystyle x^{3} = \frac{17}{4} + 3x$

$\displaystyle \Rightarrow x^{3} - 3x = \frac{17}{4}$

Answer: Shown: $\displaystyle x^{3} - 3x = \frac{17}{4}$

$= (81)^{0.6 - 0.35} = (81)^{0.25}$

$\displaystyle 0.25 = \frac{1}{4}$

$\displaystyle = (3^{4})^{\frac{1}{4}} = 3$

Answer: $3$

$\displaystyle = \left\{\frac{(2^{4})^{\frac{3}{4}}}{2^{5}}\right\}^{-\frac{1}{2}}$

$\displaystyle = \left\{\frac{2^{3}}{2^{5}}\right\}^{-\frac{1}{2}} = \left\{2^{-2}\right\}^{-\frac{1}{2}}$

$= 2^{1} = 2$

Answer: $2$