Asked by Ifechukwu

Solve for x if
I: Log2(x²-2x+5)=2
ii: Log 3(x²+2x+2)=0
Answered by Kenny

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)




Answered by David
Log3(x2+2x+2)=0
Answered by Anonymous
Bravo. Smart you
Answered by Nifemi
Hwkajaiakajan
Answered by Isaiah
Show me answer
Answered by Florence
Very good
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