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Definition of a Determinant

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Question

Show that: $2^{(\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}) . \sqrt{l o g_{a} b}}$
$= {\begin{matrix} 2, b \geq a \geq 1 2^{l o g_{a} b}, 1 < b < a \end{matrix}$

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Solution

Since $\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}}$
$= \sqrt{\frac{1}{4} l o g_{a} (a b) + \frac{1}{4} l o g_{b} (a b)}$
$= \sqrt{\frac{1}{4} (1 + l o g_{a} b + l o g_{b} a + 1} = \sqrt{⎛ ⎜ ⎝ \frac{l o g_{a} b + \frac{1}{l o g_{a} b} + 2}{4} ⎞ ⎟ ⎠}$
$= \sqrt{{⎛ ⎜ ⎜ ⎝ \frac{\sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}}}{2} ⎞ ⎟ ⎟ ⎠}^{2}}$
$= \frac{1}{2} (\sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}})$
Consider $\sqrt{l o g_{a} \sqrt[4]{(b / a)} + l o g_{b} \sqrt[4]{(a / b)}}$
$= \sqrt{\frac{1}{4} l o g_{a} (\frac{b}{a}) + \frac{1}{4} l o g_{b} (\frac{a}{b})}$
$= \sqrt{\frac{1}{4} (l o g_{a} b - 1 + l o g_{b} a - 1)} = \sqrt{\frac{l o g_{b} b + \frac{1}{l o g_{a} b} - 2}{4}}$
$= \frac{∣ ∣ ∣ \sqrt{| l o g_{a} b | - \frac{1}{\sqrt{| l o g_{a} b |}}} ∣ ∣ ∣}{2}$
Now, $P (say) = \sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}$
$P = \frac{1}{2} ⎧ ⎨ ⎩ \sqrt{| l o g_{a} b | + \frac{1}{\sqrt{| l o g_{a} b |}}} + ∣ ∣ ∣ ∣ \sqrt{| l o g_{a} b |} - \frac{1}{\sqrt{| l o g_{a} b |}} ∣ ∣ ∣ ∣ ⎫ ⎬ ⎭$
Case I : If $b \geq a \geq 1, then$
$P = \frac{1}{2} {\sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}} - \sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}}}$

$= \frac{1}{\sqrt{l o g_{a} b}}$
$∴ 2^{P \sqrt{l o g_{a} b}} = 2^{1} = 2$
Case II : If 1 < b < a, then
$P = \frac{1}{2} {\sqrt{| l o g_{a} b |} + \frac{1}{\sqrt{| l o g_{a} b |}} + \sqrt{| l o g_{a} b |} - \frac{1}{\sqrt{| l o g_{a} b |}}}$
$= \sqrt{l o g_{a} b}$
$∴ 2^{P \sqrt{l o g_{a} b}} = 2^{l o g_{a} b}$

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