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A^x=b^y=c^z ,and also a^3=b^2c,then find3/x-2/y |
Answer» ANSWER :3/x - 2/y = 1/z SOLUTION :Given :a^x = b^y = c^z a³ = b²c To find :3/x - 2/y = ? We have , a^x = b^y = c^z LET a^x = b^y = c^z = k Thus , If a^x = k , then a = k^(1/x) If b^y = k , then b = k^(1/y) If c^z = k , then c = k^(1/z) ALSO , It is given that , => a³ = b²c => [k^(1/x)]³ = [k^(1/y)]²•[k^(1/z)] => k^(3/x) = [k^(2/y)]•[k^(1/z)] => k^(3/x) = k^(2/y + 1/z) => 3/x = 2/y + 1/z => 3/x - 2/y = 1/z HENCE ,3/x - 2/y = 1/z |
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