Question

Please solve this question It's urgent please my friends help me

Answer 1

$1. \: { (\frac{5}{3} )}^{ - 2} \times {( \frac{5}{3} )}^{ - 14} = { (\frac{5}{ 3} )}^{8x} \\ \\ multiply \: the \: terms \: with \: same \: base \: by \: adding \: their \: exponents\\ \\ ( \frac{5}{3} {)}^{ - 2 - 14} = (\frac{5}{3} {)}^{8x} \\ \\ {( \frac{5}{3}) }^{ - 16} = ( \frac{5}{3} {)}^{8x} \\ \\ bases \: are \: same \: so \: set \: the \: exponents \: equal \\ \\ - 16 = 8x \\ \\ - \frac{16}{8} = x \\ \\ - 2 = x$

$2. \: ( { - 2)}^{3} \times ( - 2 {)}^{ - 6} =( - 2 {)}^{2x - 1} \\ \\ multiply \\ \\ ( - {2)}^{3 - 6} = ( - 2 {)}^{2 x - 1 } \\ \\ - 2 {}^{ - 3} = - 2 {}^{2x - 1} \\ \\ bases \: are \: same \: so \: set \: the \: exponents \: equal \\ \\ - 3 = 2x - 1 \\ \\ - 3 + 1 = 2x \\ \\ - 2 = 2x \\ \\ \frac{ - 2}{2} = x \\ \\ - 1 = x$

$3. \: ( {2}^{ - 1} + {4}^{ - 1} + {6}^{ - 1} + {8}^{ - 1} {)}^{x} = 1 \\ \\ 4 {}^{ - 1} = {2}^{ - 2} \\ {8}^{ - 1 } = {2}^{ - 3} \\ \\ ( {2}^{ - 1} + {2}^{ - 2} + {6}^{ - 1} + {2}^{ - 3) {}^{x} } = 1 \\ \\ write \: the \: no. \: in \: exponental \: form \: with \: a \: base \\ {2}^{ - 1} + {2}^{ - 2} + {6}^{ - 1} + {2}^{ - 3}$


$( 2 {}^{ - 1} + {2}^{ - 2} + 6 {}^{ - 1} + {2}^{ - 3} ) {}^{x} = (2 {}^{ - 1} + {2}^{ - 2} + {6}^{ - 1} + {2}^{ - 3} ) {}^{0} \\ \\ bases \: are \: same \: set \: the \: exponents \: equal \\ \\ x = 0$


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