Question

If log x/(a+b-2c) = log y/(b+c-2a) = log z/(c+a-2b) then the value of x²y²z² is equal to what?

Answer 1

If logx/b+c-2a=logy/c+a-2b=logz/a+b-2c, then how do you show that xyz=1?

Luckly after trying hard I got the solution. I will write down the steps below.

Let, Logx/b+c-2a=Logy/c+a-2b=Logz/a+b-2c=k

Therefore,

Logx/b+c-2a=k

Logx = k(b+c-2a) ……..(I)

Similarly ,

Logy = k(c+a-2b)…….(II)

And,

Logz = k(a+b-2c)………(III)

ADDING (I),(II)and(III)

Therefore ,

Logx + Logy + Log z = k(b+c-2a)+k(c+a-2b)+k(a+b-2c)

Log x+Logy+Logz = k(b+c-2a+c+a-2b+a+b-2c)

Log xyz = k(0)

Log xyz = 0

Log xyz = Log 1 ….. (log 1 to any base is 0)

xyz = 1

Hence, proved ……