Math, asked by zannatYasmin100, 4 months ago

please answer me..... ​

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Answered by senboni123456
2

Step-by-step explanation:

We have,

 {(yz)}^{ log(y)   - log(z) } \times  {(zx)}^{ log(z)   - log(x) } \times {(xy)}^{ log(x)   - log(y) } \\

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

 =  {(x)}^{ log( \frac{z}{x} ) +  log( \frac{x}{y} )  }  \times  {y}^{ log( \frac{y}{z} )  +  log( \frac{x}{y} ) }  \times  {z}^{ log( \frac{y}{z} )  +  log( \frac{z}{x} ) }

 =  {(x)}^{ log(z)  -  log(y) }  \times  {(y)}^{ log(x)  -  log(z) } \times  {(z)}^{ log(y) -  log(x)  }

 =  \frac{(x)^{ log(z) } }{ {(x)}^{ log(y) } }  \times  \frac{ {(y)}^{ log(x) } }{ {(y)}^{ log(z) } }  \times  \frac{ {(z)}^{ log(y) }  }{ {(z)}^{ log(x) } }

 =  \frac{(x)^{ log(z) } }{ {(y)}^{ log(x) } }  \times  \frac{ {(y)}^{ log(x) } }{ {(z)}^{ log(y) } }  \times  \frac{ {(z)}^{ log(y) }  }{ {(x)}^{ log(z) } }

 = 1

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