пожалуййстаааа помогите решить неравенство:...

Question

пожалуййстаааа помогите решить неравенство: 1)log4(2x-6)<1;</p>

2)log0,3(3x-5)<0;</p>

3)log1/3(2x-1)>-2;

4)log2(x-4)>3;

5)log2(x-7)>1;

6)log1/x(x-5)<-2</p>

Answer 1

0}} \right.\\ \left \{ {{2x-6-4<0} \atop {2x>6}} \right.\\ \left \{ {{2x<10} \atop {2x>6}} \right.\\ \left \{ {{x<5} \atop {x>3}} \right.\\ otvet (3;5)" alt="1)log4(2x-6)<1;\\ \left \{ {{(2x-6)<4^{1}} \atop {(2x-6)>0}} \right.\\ \left \{ {{2x-6-4<0} \atop {2x>6}} \right.\\ \left \{ {{2x<10} \atop {2x>6}} \right.\\ \left \{ {{x<5} \atop {x>3}} \right.\\ otvet (3;5)" align="absmiddle" class="latex-formula">

0,3^{0}\\ 3x-5>0\\ 3x>5\\ x>5/3\\ otvet\ (5/3; +beskonechnost) " alt="2)log 0,3 (3x-5)<0;\\ (3x-5)>0,3^{0}\\ 3x-5>0\\ 3x>5\\ x>5/3\\ otvet\ (5/3; +beskonechnost) " align="absmiddle" class="latex-formula">

-2;\\ \left \{ {{2x-1<(1/3)^{-2}} \atop {2x-1>0}} \right.\\ \left \{ {{2x-1<9} \atop {2x-1>0}} \right.\\ \left \{ {{2x<10} \atop {2x>1}} \right.\\ \left \{ {{x<5 \atop {x>1/2}} \right.\\ otvet\ (1/2 ;5)" alt="3)log1/3(2x-1)>-2;\\ \left \{ {{2x-1<(1/3)^{-2}} \atop {2x-1>0}} \right.\\ \left \{ {{2x-1<9} \atop {2x-1>0}} \right.\\ \left \{ {{2x<10} \atop {2x>1}} \right.\\ \left \{ {{x<5 \atop {x>1/2}} \right.\\ otvet\ (1/2 ;5)" align="absmiddle" class="latex-formula">

3;\\ x-4>2^{3}\\ x-4>8\\ x-4-8>0\\ x>12\\ otvet (12; +beskonechnost)" alt="4)log2(x-4)>3;\\ x-4>2^{3}\\ x-4>8\\ x-4-8>0\\ x>12\\ otvet (12; +beskonechnost)" align="absmiddle" class="latex-formula">

1;\\ x-7>2\\ x-7-2>0\\ x>9\\ otvet (9;+beskonechnost)" alt="5)log2(x-7)>1;\\ x-7>2\\ x-7-2>0\\ x>9\\ otvet (9;+beskonechnost)" align="absmiddle" class="latex-formula">

(1/x)^{-2}\\ x-5>x^{-2}\\ x-5-x^{-2}>0\\ -x^{-2}+x-5>0\\ x^{-2}-x+5<0\\ D=-19\\ otvet net korney" alt="6)log1/x(x-5)<-2\\ x-5>(1/x)^{-2}\\ x-5>x^{-2}\\ x-5-x^{-2}>0\\ -x^{-2}+x-5>0\\ x^{-2}-x+5<0\\ D=-19\\ otvet net korney" align="absmiddle" class="latex-formula">

 

Answer 2

 1)log4(2x-6)<1; ==> log4(2x-6)  2x-6 < 4 ==>  2x < 4+6  ==>  x < 10/2=5

2) log0,3(3x-5)<0;  ==>  log0,3(3x-5)<<span>log0,3(1)<0 ==>  3x-5 < 1  ==> 3x < 1+5 ==> x < 6/3 = 2

остальные решаешь аналогично

 3) x>5

4) x>12

5) x>9

6)log1/x(x-5)<-2  ==>  log1/x(x-5)< <span>log1/x((1/x)^(-2))  ==> log1/x(x-5)< log1/x(x^2) ==> 

==> x-5 < x^2  ==> x^2 - x +5 >0

вычисляем производную = 2х -1 

приравниваем к нулю и навходим точку минимума

2х -1 = 0  ==>  x=1/2 = 0.5

в этой точке  x^2 - x +5 = 0.5^2 - 0.5 +5 = 4.75  ==>

неравенство выполняется при всех значения х