Problem 5 - Solve the inequality

[Math Tutor]

Solve the inequality:
[tex](3-x)^{\frac{3x-5}{3-x}} < 1[/tex]

[martin123456]

Solve the inequality:
[tex](3-x)^{\frac{3x-5}{3-x}} < 1[/tex]

allowed values: x<3, x \diff 2

1)3x-5=0 =>the ineq doesnt hold
2)3x-5>0 =>ineq <=> 3-x<1 => answer is (5/3,2) and (2,3)
3)3x-5<0=>ineq <=> 3-x>1 => answer is x<5/3

finally, the answer is x<3 without x=5/3, x=2

[martin123456]

Solve the inequality:
[tex](3-x)^{\frac{3x-5}{3-x}} < 1[/tex]

allowed values: x<3, x \diff 2

1)3x-5=0 =>the ineq doesnt hold
2)3x-5>0 =>ineq <=> 3-x<1 => answer is (5/3,2) and (2,3)
3)3x-5<0=>ineq <=> 3-x>1 => answer is x<5/3

finally, the answer is x<3 without x=5/3, x=2

2) is (2,3)
so answer is x<5/3 and x in (2,3)

[Math Tutor]

Could you write the solution, please?

[martin123456]

don't u see it right above?

[anh7]

[tex](3-x)^{\frac{3x-5}{3-x }}<1[/tex] (D=R\{3})

<=>[tex](3-x)^{\frac{3x-5}{3-x }} < (3-x)^0[/tex]

_If (3-x)=1
[tex](3-x)^{\frac{3x-5}{3-x }}<1[/tex]
<=>1^3x-5<1
It can't be right



_If (3-x)>1 or (3-x)<-1 <=> x<2 or x>4

[tex](3-x)^{\frac{3x-5}{3-x }}<1[/tex]
<=>[tex]\frac{3x-5}{3-x }[/tex]<0
<=> 3x-5<0 <=> 3x<5 <=> x<[tex]\frac{5}{3 }[/tex] (*1)



_If -1<(3-x)<1 <=> 2<x<4 (x≠3)

[tex](3-x)^{\frac{3x-5}{3-x }}<1[/tex]
<=> [tex]\frac{3x-5}{3-x }[/tex]>0
<=> 3x-5>0 <=> 3x>5 <=> [tex]x>\frac{5}{3 }[/tex] (x≠3) (*2)

According (*1) and (*2) : the result will be x [tex]\in[/tex](-∞;[tex]\frac{5}{6 }[/tex]) [tex]\cup[/tex]( [tex]\frac{5}{ 3}[/tex];4)

Hope it's right

[Shpresim]

anh7, try putting a value larger than 4 not in your interval, you'll see that it will satisfy the original inequality.

I think that the solution is:

[tex](3-x)^(3x-5)/(3-x) < 1[/tex]

if a^b<c^b then b<c

[tex](3-x)^(3x-5)/(3-x) < (3-x)^0[/tex]
so

(3x-5)/(3-x)<0

the critical points are 5/3 and 3
and the solution is:

[tex]x\in[/tex] [tex](- \infty ,\frac{5}{3} )[/tex] [tex]\cup[/tex] [tex](3,+ \infty )[/tex]

[ahmed bakoush]

It's too complex to solve think and will solve in anther time.