IIT JEE , AIEEE Forum - Mathematics , Algebra

rohitkuruvila

Find the domain of ln(sin(ln(x)))

is the answer all values greater than eⁿwhere n is an integer

sankydreams

sin(ln(x)) must be positive
=> 0=> 1
i think this must be the answer.................rate if correct and coorect if wrong :-)

vivsarda

even x must nt lie between 0 n 1

apurviitjee2008

sin x has all real nos on its domain
so with respect to sin x there is no restriction
ln x is not defined for -ve nos (and 0 also)
hence with respect to ln x ,x >0
ln(sin z)
again cannot be defined for -ve nos. , so , sin z should be positive and not equal to 0 where z=ln x
hence since sin z lies bet (0,1]
so z lies between (0, (n+1)pie/2]
z=ln x lies between (0, (n+1)pie/2] so x lies between (e^0 ,e^(n+1)pie/2]

spideyunlimited

@APURV why does z lie between 0 and pi/2? it should be 0 to pi... sin is positive in 1st 2 quadrants.

SOLUTION

x > 0 for lnx

and

sin( ln(x) ) > 0 for ln [sin(ln(x)) ]
so 0< ln(x) < pi because sin is positive in 1st and 2nd quadrants.
or
e^0 < x < e^pi

1 < x < e^pi

so

x ( e⁽²ⁿ⁾, e⁽²ⁿ⁺¹⁾ ) is the general solution

priyesh

spideyunlimited a.k.a gauravragtah a.k.a artemis fowl is correct

spideyunlimited

aaaye :D ... aka aka aka hehehe///
rate kaun karega?
and apurv ka wrong tha

apurviitjee2008

SORRY FOR THE MISTAKE
I DID NOT CONSIDER THE 2ND QUADRANT WHERE SIN X HAS VALUES FROM [1,0]
SPIDEYUNLIMITED IS RIGHT......

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