# If `logx/(b-c) =logy/(c-a) = logz/(a-b)` then prove `x^a y^b z^c=1`

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### 1 Answer

The variables satisfy the relation `log x/(b-c) = logy/(c-a) = (log z)/(a-b)` .

This relation can be written as:

`log x/(b-c) = logy/(c-a) = (log z)/(a-b) = T` , where T is a constant

`log x = T*(b - c)`

`=> x = 10^(T*(b - c))`

`log y = T*(c - a)`

`=> y = 10^(T*(c - a))`

`log z = T*(a - b)`

`=> z = 10^(T*(a - b))`

Substituting for x, y and z in `x^a*y^b*z^c` with the form derived earlier gives:

`x^a*y^b*z^c`

`= (10^(T*(b - c)))^a*(10^(T*(c - a)))^b*(10^(T*(a - b)))^c`

Use the property `(x^a)^b = x^(a*b)` and `x^a*x^b = x^(a*b)`

This gives:

`= 10^(T*a*(b - c))*10^(T*b*(c - a))*10^(T*c*(a - b))`

`= 10^(T*b*a - T*c*a + T*c*b - T*a*b + T*a*c - T*b*c)`

`= 10^0`

= 1

**This proves `x^a y^b z^c=1` if **`log x/(b-c) = logy/(c-a) = (log z)/(a-b)`