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Question 1

Question 2

Решите задачу:

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

$2.\;a)\;\left(\frac12\right)^{2x-6}\leq32\\\left(2^{-1}\right)^{2x-6}\leq2^5\\2^{6-2x}\leq2^5\\6-2x\leq5\\2x\geq6-5\\2x\leq1\\x\leq\frac12\\b)\;\frac1{36}\ \textless \ 6^{x+1}\\6^{-2}\ \textless \ 6^{x+1}\\-2\ \textless \ x+1\\x\ \textgreater \ -3$

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