Question

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

Answer 1

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