If x is the solution of log 4(2 x2)+ log 4((1/8 x))= log π( tan (π/4)), then a possible solutions of the equation y2+(x+1) y+x=0, is

Question

If x is the solution of log 4(2 x2)+ log 4((1/8 x))= log π( tan (π/4)), then a possible solutions of the equation y2+(x+1) y+x=0, is (A) y=1 (B) y=-4 (C) y= log π 4 (D) y=-1

Tardigrade Admin · Accepted Answer

log 4((2 x2/8 x))=0 ⇒ log 4((x/4))=0 ⇒ (x/4)=1 ⇒ x=4 Hence, y 2+5 y +4=0 ⇒( y +4)( y +1)=0 ⇒ y =-4 or y =-1.

Q. If $x$ is the solution of $lo g_{4} (2 x^{2}) + lo g_{4} (\frac{1}{8 x}) = lo g_{π} (tan \frac{π}{4})$ , then a possible solutions of the equation $y^{2} + (x + 1) y + x = 0$ , is

$y = 1$

$y = - 4$

$y = lo g_{π} 4$

$y = - 1$

$lo g_{4} (\frac{2 x ^{2}}{8 x}) = 0 \Rightarrow lo g_{4} (\frac{x}{4}) = 0 \Rightarrow \frac{x}{4} = 1 \Rightarrow x = 4$
Hence, $y^{2} + 5 y + 4 = 0 \Rightarrow (y + 4) (y + 1) = 0 \Rightarrow y = - 4$ or $y = - 1$ .

Q. If x is the solution of log4​(2x2)+log4​(8x1​)=logπ​(tan4π​), then a possible solutions of the equation y2+(x+1)y+x=0, is

y=1

y=−4

y=logπ​4

y=−1