Question

Maths, Class 9th.

If x⁶ — y⁶ = z⁶, then prove that log$\sf _z$(x² — y²) + log[tex \sf _z[/tex](x² + y² − xy) + log$\sf _z$(x² + y² +
xy) = 6.
​

Answer 1

Answer:

x^6-y^6=z^6

Step-by-step explanation:

= (x^3)^2-(y^3)^2 =z^6

= (x^3-y^3) (x^3+y^3)=z^6

= (x-y) (x^2+xy+x^2) (x+y) (x^2-xy+y^2) =z^6

Answer 2

$\textsf{\large{\underline{Solution}:}}$

Given That:

$\rm:\longmapsto {x}^{6} - {y}^{6} = {z}^{6}$

$\rm:\longmapsto ({x}^{3})^{2} - ({y}^{3})^{2} = {z}^{6}$

Using identity a² - b² = (a + b)(a - b), we get:

$\rm:\longmapsto ({x}^{3} + {y}^{3} )( {x}^{3} - {y}^{3}) = {z}^{6}$

$\rm:\longmapsto (x + y)( {x}^{2} - xy + {y}^{2})(x - y)( {x}^{2} + xy + {y}^{2} ) = {z}^{6}$

$\rm:\longmapsto ( {x}^{2} - {y}^{2}) ( {x}^{2} - xy + {y}^{2})( {x}^{2} + xy + {y}^{2} ) = {z}^{6}$

Taking log on both sides, we get:

$\rm:\longmapsto log_{z}[( {x}^{2} - {y}^{2}) ( {x}^{2} - xy + {y}^{2})( {x}^{2} + xy + {y}^{2})]= log_{ z }{ \: z}^{6}$

We know that:

$\rm: \longmapsto log(abc..) = log(a) + log(b) + log(c) + ...$

Therefore:

$\rm:\longmapsto log_{z}[{x}^{2} - {y}^{2}] + log_{z}[{x}^{2} - xy + {y}^{2}] + log[{x}^{2} + xy + {y}^{2}]= log_{ z }{ \: z}^{6}$

$\rm:\longmapsto log_{z}[{x}^{2} - {y}^{2}] + log_{z}[{x}^{2} - xy + {y}^{2}] + log[{x}^{2} + xy + {y}^{2}]= 6 log_{ z }{z}$

$\rm:\longmapsto log_{z}[{x}^{2} - {y}^{2}] + log_{z}[{x}^{2} - xy + {y}^{2}] + log[{x}^{2} + xy + {y}^{2}]= 6$

Hence Proved..!!

$\textsf{\large{\underline{More To Know}:}}$

$\rm 1. \: \: {a}^{n} = b \implies log_{a}(b) = n$

$\rm 2. \: \: log_{a}(1) = 0, \: a \neq0,1$

$\rm 3. \: \: log_{a}(a) = 1, \: a \neq0,1$

$\rm 4. \: \: log_{a}(x) = log_{a}(y) \implies x = y$

$\rm 5. \: \: log_{e}(x) = ln(x)$

$\rm6. \: \: log_{a}(x) + log_{a}(y) = log_{a}(xy)$

$\rm7. \: \: log_{a}(x) - log_{a}(y) = log_{a} \bigg( \dfrac{x}{y} \bigg)$

$\rm 8. \: \: log_{a}( {x}^{n} ) = n\log_{a}(x)$

$\rm 9. \: \: log_{a}(m) = \dfrac{ log_{b}(m) }{ log_{b}(a) },m > 0,b > 0,a \ne1,b \ne1$

$\rm 10. \: \: log_{a}(b) = \dfrac{1}{ log_{b}(a) }$