Question

Find domain of :
log7log5log3log2(2x3+5x2-14x) ?

Answer 1

for log to be defined ,
(2x³ + 5x² -14x) > 0
x(4 x + 5 - √137)(4x + 5 +√137) > 0

log₂( 2x³ + 5x² -14x) > 0
2x³ + 5x +14x > 1
2x³ + 5x - 14x - 1 > 0________(1)

log₃log₂( 2x³ + 5x² -14x) > 0
2x³ + 5x² - 14x -2 > 0______(2)

log₅log₃log₂( 2x³ + 5x² -14x) >0
2x³ + 5x² -14x - 8 > 0_______(3)

hence,
x(4 x + 5 - √137)(4x + 5 +√137) > 0
and 2x³ + 5x² -14x - 8 > 0 after taking common (1) , (2) and (3)
(x - 2) (x + 4) (2 x + 1)>0

now put in number line then,
domain : 1/4 (-5 - sqrt(137))<x<0 or x >(-5 + sqrt(137))/4

Answer 2

Answer:for log to be defined ,

(2x³ + 5x² -14x) > 0

x(4 x + 5 - √137)(4x + 5 +√137) > 0

log₂( 2x³ + 5x² -14x) > 0

2x³ + 5x +14x > 1

2x³ + 5x - 14x - 1 > 0________(1)

log₃log₂( 2x³ + 5x² -14x) > 0

2x³ + 5x² - 14x -2 > 0______(2)

log₅log₃log₂( 2x³ + 5x² -14x) >0

2x³ + 5x² -14x - 8 > 0_______(3)

hence,

x(4 x + 5 - √137)(4x + 5 +√137) > 0

and 2x³ + 5x² -14x - 8 > 0 after taking common (1) , (2) and (3)

(x - 2) (x + 4) (2 x + 1)>0

now put in number line then,

domain : 1/4 (-5 - sqrt(137))<x<0 or x >(-5 + sqrt(137))/4

Step-by-step explanation: