# Solve this problem : If (Log x)/(b-c)=(Log y)/(c-a)=(Log z)/(a-b) Prove that x^(b+c)*y^(c+a)*z^(b+a)=1

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### 5 Answers

bullshit

We'll take logarithms both sides:

log [x^(b+c)*y^(c+a)*z^(b+a)] = log 1 (*)

We'll transform the logarithm of the product into a sum:

log [x^(b+c)*y^(c+a)*z^(b+a)] = log x^(b+c) + logy^(c+a) + log z^(b+a)

We'll use the power property of logarithms:

log x^(b+c) = (b+c)*log x

logy^(c+a) = (c+a)*log y

log z^(b+a) = (b+a)*log z

The relation (*) will become (1):

(b+c)*log x + (c+a)*log y + (b+a)*log z = 0 (1)

But, from enunciation, we'll have:

(Log x)/(b-c) = (Log y)/(c-a) = (Log z)/(a-b)

We'll cross multiply:

(Log x)*(c-a) = (Log y)*(b-c)

Log y = (Log x)*(c-a)/(b-c)

Log z = (Log x)*(a-b)/(b-c)

So, the relation (1) will become (2):

(b+c)*Log x + (c+a)*(Log x)*(c-a)/(b-c) + (b+a)*(Log x)*(a-b)/(b-c) = 0

We'll factorize by log x:

log x [b+c + (c^2-a^2)/(b-c) + (a^2 - b^2)/(b-c)] = 0

We'll divide by log x and we'll remove the brackets:

b^2 - c^2 + c^2-a^2 + a^2 - b^2 = 0

We'll remove like terms and we'll get:

**0 = 0**

**So, x^(b+c)*y^(c+a)*z^(b+a)=1 if and only if (Log x)/(b-c) = (Log y)/(c-a) = (Log z)/(a-b).**

Given (logx)/(b-c) = (logy)/(c-a) = (logz)/(a-b) . Let each equal to k.

To prove x^(b+c)*y^(c+a)*z^(a+b) = 1.

(logx)/(b-c) = k,

logx = k(b-c)

x = 10 ^k(b-c).

Therefore x^(b+c) = {10^k(b-c)}^(b+c),

x^(b+c) = 10^k(b-c)(b+c) , index law (a^m)^n = a^(mn)

x^(b+c) = 10^k(b^2-c^2)

Similarly y = 10^k(c-a) and y^(c+a) = 10^k(c^2-a^2)

z = 10^k(a-b). And so, z^(b+a) = 10^k(b+a).

Therefore x^(b+c)*y^(c+a)*z^(a+b) = 10^k{b^2-c^2+c^2-a^2+a^2-b^2} = 10^k(0) = 10^0 = 1

Given (logx)/(b-c) = (logy)/(c-a) = (logz)/(a-b) . Let each equal to k.

To prove x^(b+c)*y^(c+a)*z^(a+b) = 1.

(logx)/(b-c) = k,

logx = k(b-c)

x = 10 ^k(b-c).

Therefore x^(b+c) = {10^k(b-c)}^(b+c),

x^(b+c) = 10^k(b-c)(b+c) , index law (a^m)^n = a^(mn)

x^(b+c) = 10^k(b^2-c^2)

Similarly y = 10^k(c-a) and y^(c+a) = 10^k(c^2-a^2)

z = 10^k(a-b). And so, z^(b+a) = 10^k(b+a).

Therefore x^(b+c)*y^(c+a)*z^(a+b) = 10^k{b^2-c^2+c^2-a^2+a^2-b^2} = 10^k(0) = 10^0 = 1

let logx/b-c=logy/c-a=logz/a-b=k

therefore: multiplying logx/b-c with b+c , logy/c-a with c+a , logz/a-b with a+b.

therefore: (b+c)*logx/(b+c)(b-c)+(c-a)*logy/(c-a)(c+a)+ (a+b)*logz/(a+b)(a-b)=k

therefore: logx^(b+c)/(b^2-c^2)+logy^(c+a)/(c^2- a^2)+logz^(a+b)/(a^2-b^2)=k

therefore: log{x^(b+c)*y^(c+a)*z^(a+b)}=k*(b^2-c^2)+(c^2- a^2)+(a^2-b^2)

therefore: log{x^(b+c)*y^(c+a)*z^(a+b)}=k{b^2-c^2+c^2- a^2+a^2-b^2}

therefore: log{x^(b+c)*y^(c+a)*z^(a+b)}=k{0}

therefore: log{x^(b+c)*y^(c+a)*z^(a+b)}=0

therefore: log{x^(b+c)*y^(c+a)*z^(a+b)}=log 1 base 10

therefore: taking anti-log on both sides....

therefore: {x^(b+c)*y^(c+a)*z^(a+b)}=1

hence proved!!