Question

xlogy-logz. ylogz-logx. zlogx-logy=1

Answer 1

Given:

       $x^{log \ y - log z} \times y^{log z - log x} \times z^{log x - log y}$

To Prove:

     $x^{log y - log z} \times y^{log z - log x} \times z^{log x - log y} = 1$

Solution:

Let

     $x^{log y - log z} \times y^{log z - log x} \times z^{log x - log y} = k$

Taking log on both hand side we get

⇒ $log k = log [x^{log y - log z} \times y^{log z - log x} \times z^{log x - log y} ]$

⇒ $log k = log \ x^{log y - log z} + log \ y^{log z - log x} + log \ z^{log x - log y}$                                      

We know  $log ( a \times b ) = log \ a + log \ b$

⇒ $log k =( log y - log z) \ log (x) +(log z - log x)\ log( y) + (log\ x - log\ y) \ log ( z)$                

We know $log ( a^ b ) = b \ log a$

⇒  $log k =log x \log y - log x \ log z+ log y\ log z - log x \ log y + log x\ log z - log y\ log z$

⇒  $log k =0$

⇒   $k = e^0$          

We know    $log_a x = c \\So\ x = a^c$

⇒      $k = 1$                                                                                                                        We know    $(Anything )^0 = 1$

Therefore,

      $x^{log y - log z} \times y^{log z - log x} \times z^{log x - log y} = 1$

Hence proved