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If logXty __(log x + log y), theoglog x + logy), then show3 23that+7.ŃŃ |
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Answer» log [ ( x + y )/3] = 1/2 ( logx + logy) 2× log[ ( x + y )/3 ] = log x + log y log [ ( x + y ) / 3 ]^2 = log xy { since i ) n log a = log a^n ii ) log a + log b = log ab } Remove log both sides, [ ( x + y ) / 3 ]^2 = xy ( x + y )^2 / 3^2 = xy x^2 + y^2 + 2xy = 9xy x^2 + y^2 = 9xy - 2xy x^2 + y^2 = 7xy Divide each term with xy x^2 / xy + y^2 / xy = 7xy / xy x / y + y / x = 7 |
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