Math, asked by Divyankasc, 1 year ago

Answer this please!

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Answered by FlameFires

Ignore earlier pic. I missed a modulus sign..

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FlameFires: 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.

Divyankasc: Oh oh oh

FlameFires: 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.

Divyankasc: Yeah

Divyankasc: I'm tired of thanking now xD I never get tired but then you would be tired by listening to it xD

FlameFires: lol

abhi178: lol

abhi178: hehe

Answered by abhi178

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

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