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Answer 1

Answer:

Required result of the given equation is 9.

Step-by-step explanation:

$\implies 4^{ 5 log_{4\sqrt{2}} (3-\sqrt{6}) - 6 log_{8}(\sqrt3 - \sqrt2)}$

From the properties of logarithms :

• $log_{a} m =\dfrac{log_{b} m}{log_{b} a }$

By using the property given above :

$=> {log}_{4\sqrt2}(3-\sqrt{6})</p><p>\\\\=> \dfrac{log_{\sqrt2} (3-\sqrt6)}{log_{\sqrt2} (4\sqrt2)}\\\\\\\ => \dfrac{log_{\sqrt{2}} (3-\sqrt6)}{log_{\sqrt2}(\sqrt2)^5 }$

We know, $log_{a} a = 1$

$=> \dfrac{log_{\sqrt2}(3-\sqrt6)}{5 log_{\sqrt2}(\sqrt2)}\\\\\\\implies \dfrac{log_{\sqrt{2}}(3-\sqrt6)}{5}$

Similarly,

$=> {log}_{8}(\sqrt3 -\sqrt{2})</p><p>\\\\=> \dfrac{log_{\sqrt2} (\sqrt3-\sqrt2)}{log_{\sqrt2} (8)}\\\\\\\ => \dfrac{log_{\sqrt{2} }(\sqrt3-\sqrt2)}{log_{\sqrt2}(\sqrt2)^6 }$

We know, $log_{a} a = 1$

$=> \dfrac{log_{\sqrt2}(\sqrt3-\sqrt2)}{6 log_{\sqrt2}(\sqrt2)}\\\\\\\implies \dfrac{log_{\sqrt{2}}(\sqrt3-\sqrt2)}{6}$

Thus,

$\implies 4^{ 5\bigg(\frac{ log_{\sqrt2}(3-\sqrt6) }{5} \bigg) - 6\bigg(\frac{ log_{sqrt2} (\sqrt3 - \sqrt2)}{6} \bigg) }$

$\implies 4^{ log_{\sqrt2}(3-\sqrt6) - log_{\sqrt2}(\sqrt3-\sqrt2)}$

We know, log a - log b = log( a / b )

$\implies 4^{ log_{\sqrt2} \frac{3-\sqrt6}{\sqrt3-\sqrt2}}$

$\implies 4^{ log_{\sqrt2} \frac{\sqrt3(\sqrt3-\sqrt2)}{\sqrt3-\sqrt2}}$

$\implies 4^{ log_{\sqrt2} \sqrt3 }$

$\implies (\sqrt2)^{ 4 \times log_{\sqrt2} \sqrt3}$

Using, n log b = log bⁿ

$\implies (\sqrt2)^{ log_{\sqrt2}(\sqrt3)^4}$

Using, $a^{log_{a} b} = b$

$\implies (\sqrt3)^4$

= > 9

Hence the required result of the given equation is 9.

Answer 2

Step-by-step explanation:

Hope it helps you Tuank