Question

Log7(5+4x-x^)+logx-1 3

Answer 1

1)ОДЗ x>0 U x≠1⇒x∈(0;1) U (1;∞)log(7)x- 1=6/log(7)x
(log(7)x)²- log(7)x-6=0
log(7)x=a
a²-a-6=0
a1+a2=1 U a1*a2=-6
a1=-2⇒log(7)x=-2⇒x=1/49
a2=3⇒log(7)x=3⇒x=343
2)ОДЗ x>0 U x≠1 U x≠1/2 U x≠1/4⇒x∈(0;1/4) U (1/4;1/2) U (1/2;1)U (1;∞)
1/log(2)x *1/(log(2)x + 1)=1/(log(2) +2)
log(2)x*(log(2)x +1)=log(2)x+2
(log(2)x)²+log(2)x-log(2)x=2
(log(2)x)²=2
log(2)x=-√2⇒x=1/2 ^{ \sqrt{2} }
log(2)x=√2⇒x=2 ^{ \sqrt{2} }
3)ОДЗ x>0
0,4 ^{lg^2x-1} =0,4 ^{2lgx^2+4}
lg²x-1=4+4lgx
lg²x-4lgx-5=0
lgx=a
a²-4a-5=0
a1+a2=4 U a1*a2=-5
a1=-1⇒lgx=-1⇒x=0,1
a2=5⇒lgx=5⇒x=100000

Answer 2

It appears that you have an expression involving logarithms with different bases, variables, and constants. Let's break down the expression step by step:

The expression you provided is: log7(5 + 4x - x^) + logx - 1 3

However, there seems to be a formatting issue or missing information in your expression. The term "x^" is not properly defined, and there's a "-1 3" at the end of the expression that doesn't appear to belong there.

If you could provide a properly formatted expression or clarify what you intended, I would be happy to help you further.