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Basic of Logarithms
hard
Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$
A
$A = 1$
B
$B = -3$
C
$C = 2$
D
All of these
Solution
(d) ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}} = {{7\,.\,({2^{1/4}} – 1)} \over {({2^{1/4}} – 1)\,[{{({2^{1/4}})}^2} + {2^{1/4}}.1 + {1^2}]}}$
$= {{7\,.\,({2^{1/4}} – 1)} \over {{2^{3/4}} – 1}} = A + B\,.\,{2^{1/4}} + C.\,{2^{1/2}} + D{.2^{3/4}}$
==> $7\,.\,{2^{1/4}} – 7 = (A – D)\,{2^{3/4}} + (2B – A) + (2C – B){.2^{1/4}}$$ + (2D – C){2^{1/2}}$
==> $(2B – A + 7) + (A – D){2^{3/4}} + (2C – B – 7){2^{1/4}}$ $+ (2D – C){2^{1/2}} = 0$
==> $2B – A + 7 = A – D = 2C – B – 7 = 2D – C = 0$
==> $A = D = 1,\,B = – 3,\,C = 2$.
Standard 11
Mathematics
