Question

if (xy)^(a-1)=z, (yz)^(b-1)=x, (xz)^(c-1)=y, and xyz is not -1,0 or 1, then which of the following is equal to ab+bc+ca?
a. abc
b. abc÷2
c. 2abc
d. 3abc
​

Answer 1

Given:

• $(xy)^{a-1}=z,\:(yz)^{b-1}=x,\:(zx)^{c-1}=y$

• $xyz\neq -1,\:0,\:1$

To find: the value of $ab+bc+ca$

Solution:

Given, $(xy)^{a-1}=z,\:(yz)^{b-1}=x,\:(zx)^{c-1}=y$

Taking $log,$ we get

• $\quad (a-1)\:log(xy)=log(z)$
• $\Rightarrow a=\frac{log(xyz)}{log(xy)}$

• $\quad (b-1)\:log(yz)=log(x)$
• $\Rightarrow b=\frac{log(xyz)}{log(yz)}$

• $\quad (c-1)\:log(zx)=log(y)$
• $\Rightarrow c=\frac{log(xyz)}{log(zx)}$

Then, $ab+bc+ca$

$=\frac{\{log(xyz)\}^{2}}{log(xy).log(yz)}+\frac{\{log(xyz)\}^{2}}{log(yz).log(zx)}+\frac{\{log(xyz)\}^{2}}{log(zx).log(xy)}$

$=\{log(xyz)\}^{2}\left[\frac{log(xy)+log(yz)+log(zx)}{log(xy).log(yz).log(zx)}\right]$

$=\{log(xyz)\}^{2}\times\frac{log(x^{2}y^{2}z^{2})}{log(xy).log(yz).log(zx)}$

$=\{log(xyz)\}^{2}\times\frac{log\{(xyz)^{2}\}}{log(xy).log(yz).log(zx)}$

$=\{log(xyz)\}^{2}\times \frac{2\times log(xyz)}{log(xy).log(yz).log(zx)}$

$=2\times\frac{\{log(xyz)\}^{3}}{log(xy).log(yz).log(zx)}$

$=2\times\frac{log(xyz)}{log(xy)}\times\frac{log(xyz)}{log(yz)}\times\frac{log(xyz)}{log(zx)}$

$=2abc$

$\Rightarrow ab+bc+ca=2abc$

Answer: Option C) 2abc is correct.