+0  
 
0
773
6
avatar+386 

 

log (x^2-4x+1)

 Dec 17, 2016

Best Answer 

 #4
avatar+37146 
+5

Here is the parabola

 Dec 17, 2016
edited by ElectricPavlov  Dec 18, 2016
 #1
avatar+37146 
0

This is not an EQUATION...this is a RELATION.

to solve for x, you need to have an = 'something'    to make it ann equation.     Sorry

 Dec 17, 2016
 #2
avatar
0

Solve for x:
(log(x^2 - 4 x + 1))/(log(10)) = 0

Multiply both sides by log(10):
log(x^2 - 4 x + 1) = 0

Cancel logarithms by taking exp of both sides:
x^2 - 4 x + 1 = 1

Subtract 1 from both sides:
x^2 - 4 x = 0

Factor x from the left hand side:
x (x - 4) = 0

Split into two equations:
x - 4 = 0 or x = 0

Add 4 to both sides:
Answer: |x = 4               or                  x = 0

 Dec 17, 2016
 #3
avatar+37146 
+5

Alright, I re-read your question....  for this to 'BE DEFINED'  ,   the value of

x^2 - 4x+1     HAS to be POSITIVE (greater than 0-- because there is no logs of NEGATIVE numbers)

so

x^2 - 4x +1 > 0

Quadratic formula results in  ZEROES at   x = 2 +-sqrt3

If you graph this fxn you will see a parabola which is POSITIVE outside of these values and NEGATIVE between them

so       2+sqrt3 <   x < 2-sqrt3

 Dec 17, 2016
 #5
avatar+386 
0

Huge thanks for your help!

 

Only thing:

 

wouldn't it be 

 

2+sqrt3 "greater or equal to" x "lesser or equal to" 2-sqrt3

 

 2+sqrt3 <   x < 2-sqrt3

 

???

TonyDrummer2  Dec 18, 2016
 #4
avatar+37146 
+5
Best Answer

Here is the parabola

ElectricPavlov Dec 17, 2016
edited by ElectricPavlov  Dec 18, 2016
 #6
avatar+129852 
+5

Note, Tony......we are interested in the values of the parabola that are greater than 0......another way of denoting this is :

 

(−inf, 2 - √3 )  U ( 2 + √3 , inf )

 

 

 

cool cool cool

 Dec 18, 2016

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