Answer this please!
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Ignore earlier pic. I missed a modulus sign..
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The property of square is to make all negative values as positive. By removing it using logarithmic property, we must retain a modulus sign so that we preserve the negative turning into positive feature of the square, or else the function would be changed.
Oh oh oh
also I made a mistake writing - 5/2 is outside range. It actually isn't. I meant to write that it will also make base if log negative which is not defined.
Yeah
I'm tired of thanking now xD I never get tired but then you would be tired by listening to it xD
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log { (x² + x -6)² base (x +1 ) } = 4
(x² + x - 6)²= (x +1)⁴
(x² + x -6)² - ( x +1)⁴ = 0
{x² + x -6 -x² -2x -1 }{ x² + x -6 +x² +2x +1 } =0
( -x -7 )( 2x² + 3x -5 ) = 0
x = -7 and 2x² + 3x -5 = 0
2x² +3x -5 = 0
2x² -2x +5x - 5= 0
(x -1)(2x +5 ) = 0
x = -5/2 , 1
now for log to be defined when ,
x + 1 ≠ 1 and x+1 > 0
x ≠ 0 and x > -1
and (x² + x -6)² > 0 for all real value of x
so , for this condition , x = 1 is possible .
so, x = 1
(x² + x - 6)²= (x +1)⁴
(x² + x -6)² - ( x +1)⁴ = 0
{x² + x -6 -x² -2x -1 }{ x² + x -6 +x² +2x +1 } =0
( -x -7 )( 2x² + 3x -5 ) = 0
x = -7 and 2x² + 3x -5 = 0
2x² +3x -5 = 0
2x² -2x +5x - 5= 0
(x -1)(2x +5 ) = 0
x = -5/2 , 1
now for log to be defined when ,
x + 1 ≠ 1 and x+1 > 0
x ≠ 0 and x > -1
and (x² + x -6)² > 0 for all real value of x
so , for this condition , x = 1 is possible .
so, x = 1
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