\( \Large 17^{3.5} \times 17^{7.3} \div 17^{4.2} = 17^{?} \) => \( \Large 17^{3.5 + 7.3 - 4.2} = 17^{?} \) => \( \Large 17^{6.6} = 17^{?} \) Therefore ? = 6.6
\( \Large ? = \sqrt{ \left(13\right)^{4} } = \sqrt{13 \times 13 \times 13 \times 13} \) \( \Large 13 \times 13 = 169 \)
Given that, \( \Large 289 = 17^{\frac{x}{5}} \) => \( \Large 17^{2} = 17^{\frac{x}{5}} => \frac{x}{5} \) =2 => x = 10
Given that, \( \Large a^{2x+2} = 1 \) => \( \Large a^{2x+2} = a^{0} \) => \( \Large 2x + 2 = 0 \) => \( \Large x = \frac{-2}{2} = -1 \)
\( \Large ? = \frac{\left[ \left(12\right)^{-2} \right]^{2}}{\left[ \left(12\right)^{2} \right]^{-2}} = \frac{ \left(12\right)^{-4} }{ \left(12\right)^{-4} }= 1 \)
\( \Large 81^{2.5} \times 9^{4.5} \div 3^{4.8} = 9^{?} \) => \( \Large \frac{81^{2.5} \times 9^{4.5}}{3^{4.8}} = 9^{?} \) => \( \Large \frac{ \left(3^{4}\right)^{2.5} \times \left(3^{2}\right)^{4.5} }{3^{4.8}} = 9^{?} \) => \( \Large \frac{3^{10} \times 3^{9}}{3^{4.8}} = 9^{?} \) => \( \Large 3^{10+9-4.8} = 9^{?} \) => \( \Large 9^{?} = 3^{14.2} \) => \( \Large \left(3\right)^{2 \times ?} = 3^{14.2} \) => \( \Large 2 \times ? = 14.2 \) Therefore, \( \Large ? = \frac{14.2}{2} = 7.1 \)
Given that, \( \Large \left(2^{4 \times \frac{1}{2}}\right)^{?} = 256 \) => \( \Large \left(2^{2}\right)^{?} = 2^{8} => 2 \times ? = 8 \) Therefore, \( \Large ? = \frac{8}{2} = 4 \)
\( \Large ? = \sqrt[3]{103823} = \sqrt[3]{ \left(47\right)^{3} } = \left[ \left(47\right)^{3} \right]^{\frac{1}{3}} \) \( \Large = 47^{3 \times \frac{1}{3}} = 47 \)
\( \Large \left(10\right)^{200} \div \left(10\right)^{196} = \left(10\right)^{200 - 196} \) = \( \Large 10^{4} = 10000 \)
Given. \( \Large xyz =1, a^{x} = b, b^{y} = c \) Now, \( \Large b = a^{x} \) => \( \Large b^{y} = a^{xy} \) => \( \Large b^{yz} = a^{xyz} \) => \( \Large c^{z} = a \)