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If log4 + log(17x+1) = log(3x+7) - 0^(0!) + 0^(0) + 9 Mod 11 - 243 Mod 13 , then value of x is ?

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If log4 + log(17x+1) = log(3x+7) - 0^(0!) + 0^(0) + 9 Mod 11 - 243 Mod 13 , then value of x is ?
posted Apr 5, 2017 by anonymous

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1 Answer

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x=1.041247503
LOG IS NATURAL LOG with LOG10 there is no solution
LN(4)+LN(17x+1)=LN(3x+7)+11-9
SIMPLIFIES TO
LN(4)+LN(17x+1)=LN(3x+7)+2 . .............(1) Taking 2 to be LN(e^2) we have
LN(17x+1)-LN(3x+7)=LN(e^2)-LN(4)
or LN((17x+1)/(3x+7))=LN((e^2)/4) hence
(17x+1)/(3x+7)=(e^2)/4 solution gives x=1.041247503

substituting this value of x into function y= LN(4)+LN(17x+1)-LN(3x+7)-2 gives y=-9.50891800410858E-07 which is as good as zero

answer Apr 6, 2017 by Kewal Panesar



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