a^x=b^y=c^z ,and also a^3=b^2c,then find3/x-2/y
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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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