Plz any bhaiya solve this..... y is no one replying??????????..................
if f(x)=log_{e} x and g(x)=x^{2} and 4<c<5
then
c log (4^{25}/5^{16}) is equal to
(a) c log_{e}5  8
(b) 2(c^{2} log_{e}4  8)
(c) 2(c^{2} log_{e}5  8)
(d) c log_{e}4  8

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8 Answers
if f(x)=loge x and g(x)=x2 and c (4,5)
wat is c(4,5) could u please clarify?
No one here who can solve this Problem.....................???????????????
rajat what is the source of this question...??
I feel something wrong or mistyped..
I dont find any use of c lying betweeen 4 and 5!!!
No bhaiya da question is correct.... it is from arihant mathematics..
look, first of all, i feel this q. is too ambiguous,
as far as the solution is concerned,
say h(x)=x^{2}log416logx
h(x) is diff. and continuous in (4,5)
we have by LMVT,
h(5)h(4) /54 = h'(c) where c is in (4,5)
so,
we have [25log4  16log5][0] = 2clog416/c
or log(4^{25}/5^{16}) =(2c^{2}log4 16)/c^{2}
that way im getting
c log(4^{25}/5^{16})= 2(c^{2}log48)/c
but i thing i can say is, the question shud ask us to make use of these functions, other wise there mite be many many ways of solving this one, which may use many different 'c' s..
cheers!!