Question

plz help me guys.....​

Answer 1

Proof :

$x=1+log_{a}bc$

$\implies x=1+\frac{log_{e}bc}{log_{e}a}$

$\implies x=\frac{log_{e}a+log_{e}bc}{log_{e}a}$

$\implies x=\frac{log_{e}abc}{log_{e}a}$

$y=1+log_{b}ca$

$\implies y=1+\frac{log_{e}ca}{log_{e}b}$

$\implies y=\frac{log_{e}b+log_{e}ca}{log_{e}b}$

$\implies y=\frac{log_{e}abc}{log_{e}b}$

$z=1+log_{c}ab$

$\implies z=1+\frac{log_{e}ab}{log_{e}c}$

$\implies z=\frac{log_{e}c+log_{e}ab}{log_{e}c}$

$\implies z=\frac{log_{e}abc}{log_{e}c}$

L.H.S. = xy + yz + zx

$=\frac{(log_{e}abc)^{2}}{log_{e}a\:log_{e}b}+\frac{(log_{e}abc)^{2}}{log_{e}b\:log_{e}c}+\frac{(log_{e}abc)^{2}}{log_{e}c\:log_{e}a}$

$=\frac{\{(log_{e}abc)^{2}\}(log_{e}c+log_{e}a+log_{e}b)}{log_{e}a\:log_{e}b\:log_{e}c}$

$=\frac{(log_{e}abc)^{3}}{log_{e}a\:log_{e}b\:log_{e}c}$

$=\frac{log_{e}abc}{log_{e}a}\frac{log_{e}abc}{log_{e}b}\frac{log_{e}abc}{log_{e}c}$

= xyz = R.H.S.

Hence, proved.

Answer 2

Answer:

Step-by-step explanation:

Proof :

x=1+log_{a}bc

\implies x=1+\frac{log_{e}bc}{log_{e}a}

\implies x=\frac{log_{e}a+log_{e}bc}{log_{e}a}

\implies x=\frac{log_{e}abc}{log_{e}a}

y=1+log_{b}ca

\implies y=1+\frac{log_{e}ca}{log_{e}b}

\implies y=\frac{log_{e}b+log_{e}ca}{log_{e}b}

\implies y=\frac{log_{e}abc}{log_{e}b}

z=1+log_{c}ab

\implies z=1+\frac{log_{e}ab}{log_{e}c}

\implies z=\frac{log_{e}c+log_{e}ab}{log_{e}c}

\implies z=\frac{log_{e}abc}{log_{e}c}

L.H.S. = xy + yz + zx

=\frac{(log_{e}abc)^{2}}{log_{e}a\:log_{e}b}+\frac{(log_{e}abc)^{2}}{log_{e}b\:log_{e}c}+\frac{(log_{e}abc)^{2}}{log_{e}c\:log_{e}a}

=\frac{\{(log_{e}abc)^{2}\}(log_{e}c+log_{e}a+log_{e}b)}{log_{e}a\:log_{e}b\:log_{e}c}

=\frac{(log_{e}abc)^{3}}{log_{e}a\:log_{e}b\:log_{e}c}

=\frac{log_{e}abc}{log_{e}a}\frac{log_{e}abc}{log_{e}b}\frac{log_{e}abc}{log_{e}c}

= xyz = R.H.S.