Question

show that log (x^2/yz) + log(y^2/zx) + log (z^2/xy) = 0​

Answer 1

Step-by-step explanation:

= log(x²/yz) + log(y²/zx) + log(z²/xy)

= log( (x²*y²*z²)/(yz*zx*xy) )

= log( (x²y²z²)/(x²y²z²) )

= log(1)

= 0

Answer 2

Question :

Show that

$\sf \log( \frac{x {}^{2} }{yz} ) + \log( \frac{y {}^{2} }{zx} ) + \log( \frac{z {}^{2} }{xy}) = 0$

Formula's Used :

$\sf1) \log x + \log \: y = \log \:xy$

$\sf2) \log( \frac{x}{y}) = \log \: x - \log \: y$

$\sf3) \log \: x {}^{n} = n \log \:x$

Solution :

LHS

$\sf = log( \frac{x {}^{2} }{yz} ) + log( \frac{y {}^{2} }{zx} ) + log( \frac{z {}^{2} }{xy} )$

We know that

$\sf \log (\frac{x}{y}) = \log \: x - \log \: y$

$\sf = log(x {}^{2} ) - log(yz) + log(y {}^{2} ) - log(zx) + log(z {}^{2} ) - log(xy)$

$\sf = logx {}^{2} + logy {}^{2} + logz{}^{2} - ( \log \: xy + \log \: yz + \log \: zx)$

Now use property

log a + log b = log ab

$\sf = \log(x {}^{2} y {}^{2} z {}^{2} ) - \log(xy \times yz \times zx)$

$\sf = \log(x {}^{2}y {}^{2} z {}^{2}) - \log(x {}^{2} y {}^{2} z {}^{2} )$

$\sf = 0$

RHS = 0

Thus LHS = RHS

Hence proved