V20 13. If (x,y,) and (x,y,) are the solution of the system of equation. log225(x) + logo(y) = 4 log, (225) – log, (64) = 1, then show that the value of log20(x,y,x,y) = 12.

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( log _{operatorname{das}} x+operatorname{cog}_{6} 4 y=4 ) ( operatorname{cog}_{x} 225-operatorname{cog}_{y} 64=1 ) ( mid log _{a} b=frac{1}{log _{b} a} ) cet ( log _{225} x=A quad & quad log _{64} y=B ) now ( quad A+B=4 quad Rightarrow B=4-A ) ( frac{1}{A}-frac{1}{B}=1 quad ) (put value of ( frac{1}{A}-frac{1}{4-A}=1 ) ( frac{4-2 A}{4 A-A^{2}}=1 quad Rightarrow quad begin{array}{l}A^{2}-6 A+4=0 text { Sridharacharya }end{array} ) ( Rightarrow A=frac{6 pm 2 sqrt{5}}{2} Rightarrow frac{sqrt{A}=3 pm sqrt{5}}{B=4-A=4-(3 pm sqrt{5})} 4 sqrt{B=1 pm sqrt{5}} ) ( log _{235} x=A=3 pm sqrt{5}left(operatorname{lct} tan 0, x, operatorname{and} x_{2}right) ) ( log _{225} x_{1}+operatorname{cog}_{225} x_{2}=3+85+3-frac{12}{3}=6=log _{225} x_{1} x_{2} ) similarly ( x_{1} x_{2}=left(15^{2}right)^{6}=15^{12} ) ( log _{64} y_{1}+log _{69} y_{2}=1+y_{5}+1-sqrt{5}=1+1=2-log _{64} y_{1} y_{2} ) ( y, y_{2}=(64)^{2}=left(2^{8}right)^{32}=2^{12} ) ( left[left(x, x_{2}right)left(y, y_{2}right)=15^{12} cdot 2^{12}=30^{12}right. )

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