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Question: Solution of the equation \(= \frac{1}{4}\left( \frac{3}{8}x + \frac{3}{4} \right)\left( 1 + \frac{x^...

Solution of the equation =14(38x+34)(1+x24)1+581(2)(1x2)1= \frac{1}{4}\left( \frac{3}{8}x + \frac{3}{4} \right)\left( 1 + \frac{x^{2}}{4} \right)^{- 1} + \frac{5}{8}\frac{1}{( - 2)}\left( 1 - \frac{x}{2} \right)^{- 1}are.

A

0

B

6

C

4

D

None of these

Answer

6

Explanation

Solution

=202log20(1/9)=202log209=20log2092=92=81[logba.logcalogaa]+[logab.logcblogbb]+[logaclogbclogcc]=0= 20^{- 2\log_{20}(1/9)} = 20^{2\log_{20}9} = 20^{\log_{20}9^{2}} = 9^{2} = 81\lbrack\log_{b}a.\log_{c}a - \log_{a}a\rbrack + \lbrack\log_{a}b.\log_{c}b - \log_{b}b\rbrack + \lbrack\log_{a}c\log_{b}c - \log_{c}c\rbrack = 0

x=94x = \frac{9}{4} 1lna.lnb.lnc[(lna)3+(lnb)3+(lnc)33lna.lnb.lnc]=0\frac{1}{\ln a.\ln b.\ln c}\lbrack(\ln a)^{3} + (\ln b)^{3} + (\ln c)^{3} - 3\ln a.\ln b.\ln c\rbrack = 0

\Rightarrow lna+lnb+lnc=0\ln a + \ln b + \ln c = 0 ln(abc)\ln(abc) [a3+b3+c33abc=0\lbrack a^{3} + b^{3} + c^{3} - 3abc = 0 a+b+c=0]a + b + c = 0\rbrack \therefore

This value satisfies the given equation.

abc=1abc = 1.