Question

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Answer 1

$1.;a);0,3^{3x-2}=1\0,3^{3x-2}=0,3^0\3x-2=0\3x=2\x=frac23\b);2^{3x+2}-2^{3x-2}=30\4cdot2^{3x}-frac{2^{3x}}4=30\16cdot2^{3x}-2^{3x}=120\15cdot2^{3x}=120\2^{3x}=8\2^{3x}=2^3\3x=3\x=1 \c);64^x-8^x-56=0\8^{2x}-8^x-56=0\8^x=t,;8^{2x}=t^2,;t textgreater 0\t^2-t-56=0\D=1+4cdot56=225=(15)^2\t_{1,2}=frac{1pm15}2\t_1=8\t_2=-7;He;nogx.\8^x=8Rightarrow x=1$

$2.;a);left(frac12right)^{2x-6}leq32\left(2^{-1}right)^{2x-6}leq2^5\2^{6-2x}leq2^5\6-2xleq5\2xgeq6-5\2xleq1\xleqfrac12\b);frac1{36} textless 6^{x+1}\6^{-2} textless 6^{x+1}\-2 textless x+1\x textgreater -3$

$3.;a)\begin{cases}2^x+2^{y+1}=10\y-x=1end{cases}Rightarrowbegin{cases}2^x+2^{x+1+1}=10\y=x+1end{cases}\2^x+2^{x+1+1}=10\2^x+2^{x+2}=10\2^x+4cdot2^x=10\5cdot2^x=10\2^x=2\x=1\begin{cases}x=1\y=2end{cases}$
$b)\begin{cases}4^xcdot2^y=32\2^{2x}-2^y=14end{cases}\4^xcdot2^y=32\2^{2x}cdot2^y=2^5\2^{2x+y}=2^5\2x+y=5\y=5-2x\begin{cases}y=5-2x\2^{2x}-2^{5-2x}=14end{cases}\2^{2x}-2^{5-2x}=14\2^{2x}-frac{32}{2^{2x}}=14;;;;times2^{2x}\2^{4x}-32=14cdot2^{2x}\2^{4x}-14cdot2^{2x}-32=0\2^{2x}=t,;2^{4x}=t^2,;t textgreater 0\t^2-14t-32=0\D=196+4cdot32=324=(18)^2\t_{1,2}=frac{14pm18}2\t_1=16\t_2=-2;-;He;nogx.\2^{2x}=16\2^{2x}=2^4\2x=4\x=2\begin{cases}y=1\x=2end{cases}$