x = 1.6(1)
10x = 16.(1)
100x = 161.(1)
100x - 10x = 161.(1) - 16.(1)
90x = 145
x = 145/90 = 29/18
Log₂(1-3x)≤4 ОДЗ: 1-3x>0 3x<1 x<1/3
log₂(1-3x)≤log₂16
1-3x≤16
-3x≤15 |÷(-3)
x≥-5 ⇒
Ответ: x∈[-5;1/3).
![2 ^{x} +80\cdot 2^{4-x} \geq 261 \\ 2 ^{x} +80\cdot 2^{4}\cdot 2 ^{-x} \geq 261 \\](https://tex.z-dn.net/?f=2+%5E%7Bx%7D+%2B80%5Ccdot+2%5E%7B4-x%7D+%5Cgeq+261+%5C%5C+2+%5E%7Bx%7D+%2B80%5Ccdot+2%5E%7B4%7D%5Ccdot+2+%5E%7B-x%7D+%5Cgeq+261+%5C%5C+)
Замена переменной
![2 ^{x}=t \\ \\ 2 ^{x}= \frac{1}{t} ,t>0](https://tex.z-dn.net/?f=2+%5E%7Bx%7D%3Dt+%5C%5C++%5C%5C+2+%5E%7Bx%7D%3D+%5Cfrac%7B1%7D%7Bt%7D+%2Ct%3E0+)
![t+ \frac{1280}{t} \geq 261 \\ \\ t ^{2} -261t=1280 \geq 0,t>0](https://tex.z-dn.net/?f=t%2B+%5Cfrac%7B1280%7D%7Bt%7D+%5Cgeq+261+%5C%5C++%5C%5C+t+%5E%7B2%7D+-261t%3D1280+%5Cgeq+0%2Ct%3E0+)
Решаем уравнение
t²-261 t+1280=0
D=261²-4·1280=68121-5120=63001=251²
t=(261-251)/2=5 или t=(261+251)/2=256
+ - +
-----------------------[5]-----------------------[256]-----------------
0<t≤5 или t≥256
![2 ^{x} \geq 256](https://tex.z-dn.net/?f=2+%5E%7Bx%7D+%5Cgeq+256+)
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![x \leq log_25](https://tex.z-dn.net/?f=x+%5Cleq+log_25)
x≥8