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Mathematics 7 Online
OpenStudy (anonymous):

what is the method of finding domain and range of Log functions like ln(2x-1)

OpenStudy (anonymous):

well, 2x-1 shud not be equal to zero greater than zero and range is R

OpenStudy (anonymous):

other than zero all the real number are included then...???

OpenStudy (anonymous):

for what? range or domain

OpenStudy (anonymous):

for range

OpenStudy (anonymous):

zero is inc in range as x=1

OpenStudy (anonymous):

i.e. 2x-1=1

OpenStudy (anonymous):

sorry i am not getting it

OpenStudy (anonymous):

log 1=0

OpenStudy (anonymous):

then domain is from 1-+infinity

OpenStudy (anonymous):

1to +infinity right..??

OpenStudy (anonymous):

nope log 1/x = -log x so domain is R

OpenStudy (anonymous):

ahan oka got it ... mean range will include zero as well as it is set w.. an domain will ve all real num xcpt zero.. right maha/?

OpenStudy (anonymous):

dom = [1,infnty

OpenStudy (anonymous):

range is for dependent variable right..??

OpenStudy (anonymous):

oops i dint read ur comment properly...domain is ok from ur side @Maha_Chauhan

OpenStudy (anonymous):

dom is from 1_+infinity and range is all real #'s rite?

OpenStudy (ranga):

You cannot take ln of anything <= 0. So 2x - 1 > 0 x > 1/2. So the domain is (1/2, infinity) The range is (-infinity, infinity). y approaces -infinity when x = 1/2 and y approaches +infinity when x approaches +infinity

OpenStudy (anonymous):

how it can be 1/2..???

OpenStudy (anonymous):

the point that i got is log's cant be of 0 and negative numbers.. right...??? @divu.mkr

OpenStudy (goformit100):

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OpenStudy (anonymous):

yep inside log it can't be zero or negative

OpenStudy (anonymous):

then it becomes domain right..??

OpenStudy (anonymous):

hmm

OpenStudy (anonymous):

okay thanku so much

OpenStudy (anonymous):

np

OpenStudy (anonymous):

in a log if the function is \[x^{2}+y^{2}\how the domain will be obtained..??

OpenStudy (ranga):

log(A), ln(A) are all defined only when A > 0 Is the question, What is the domain of \[\Large \log(x ^{2} + y ^{2})\]?

OpenStudy (anonymous):

yes @ranga

OpenStudy (ranga):

log(A) is defined ONLY for A > 0 So (x^2 + y^2) must be > 0. But both x and y are squared. So whether they x and y are positive or negative x^2 and y^2 will always be > 0. The only exception is when BOTH x and y are 0. Then x^2 + y^2 will be 0 which is not defined for log function. x can be separately 0, y can be separately 0 but they both cannot be simultaneously 0. So the domain is: x can be any real number including 0 but BOTH x and y cannot be simultaneously be 0.

OpenStudy (anonymous):

thank you...@ranga

OpenStudy (ranga):

you are welcome.

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