I assume the base is set to base 2
Log2(x²-2x+5)=2
(x²-2x+5)=2²
x²-2x+5=4
x²-2x+1=0
(x-1)²=0
x=1
Similarly
Am assuming the base is 3
Log 3(x²+2x+2)=0
(x²+2x+2)=3⁰
x²+2x+2=1
x²+2x+1=0
Factored as
(x+1)²=0
x=-1
If they are not to base 10
(x²-2x+5)=10²/2
(x²-2x+5)=50
x²-2x=45
(x-1)²=45+1
(x-1)²=46
not too nice
Same goes for
Log 3(x²+2x+2)=0
x²+2x+2=1/3
x²+2x=1/3-6/3
(x+1)²=-5/3+3/3
(x+1)²=-2/3
Which does work well we have an imaginary result
(x+1)=±i√(2/3)
Solve for x if
I: Log2(x²-2x+5)=2
ii: Log 3(x²+2x+2)=0
6 answers
Log3(x2+2x+2)=0
Bravo. Smart you
Hwkajaiakajan
Show me answer
Very good