Question

Find the value of a^logb- logc × b^logc- loga × c^loga-logb

Answer 1

Please see the attachment

Answer 2

Answer:

Value = 1

Step-by-step explanation:

As per the question,

According to logarithmic property,

$logx-logy=log(\frac{x}{y})\\logx+logy=log(xy)$

Now,

$a^{(logb-logc)}\times b^{(logc-loga)}\times c^{(loga-logb)}$

$y=a^{(logb-logc)}\times b^{(logc-loga)}\times c^{(loga-logb)}$

Taking both side log, we get

$logy=log(a^{(logb-logc)}\times b^{(logc-loga)}\times c^{(loga-logb)})$

After using property and solving we get,

$logy=loga^{(logb-logc)}+log b^{(logc-loga)}+logc^{(loga-logb)})$

$logy=(logb-logc)loga+(logc-loga)logb+(loga-logb)logc$

$logy=(logb\times loga-logc\times loga)+(logc\times logb-loga\times logb)+(loga\times logc-logb\times logc)$

As all the terms get cancelled and we get log y = 0

Therefore,

log y = 0

Which is possible only if y = 1

Hence, the required value  = 1.