Question

1÷logxy(xyz)+1÷logyz(xyz)+1÷logzx(xyz)=2​

Answer 1

Solution :

L.H.S. = $\mathrm{\frac{1}{log_{xy}(xyz)}+\frac{1}{log_{yz}(xyz)}+\frac{1}{log_{zx}(xyz)}}$

$\displaystyle\mathrm{=\frac{1}{\frac{log_{e}(xyz)}{log_{e}(xy)}}+\frac{1}{\frac{log_{e}(xyz)}{log_{e}(yz)}}+\frac{1}{\frac{log_{e}(xyz)}{log_{e}(zx)}}}$

$\displaystyle\mathrm{=\frac{log_{e}(xy)}{log_{e}(xyz)}+\frac{log_{e}(yz)}{log_{e}(xyz)}+\frac{log_{e}(zx)}{log_{e}(xyz)}}$

$\displaystyle\mathrm{=\frac{log_{e}(xy)+log_{e}(yz)+log_{e}(zx)}{log_{e}(xyz)}}$

$\displaystyle\mathrm{=\frac{log_{e}(xy\times yz\times zx)}{log_{e}(xyz)}}$

$\displaystyle\mathrm{=\frac{log_{e}(x^{2}y^{2}z^{2})}{log_{e}(xyz)}}$

$\displaystyle\mathrm{=\frac{log_{e}(xyz)^{2}}{log_{e}(xyz)}}$

$\displaystyle\mathrm{=\frac{2\:log_{e}(xyz)}{log_{e}(xyz)}}$

= 2 = R.H.S.

Hence, proved.