First slide

Trigonometric Identities

Question

Consider $E = 2^{(\sqrt{\log_{a} \sqrt[4]{ab} + \log_{b} \sqrt[4]{ab}} - \sqrt{\log_{a} \sqrt[4]{\frac{b}{a}} + \log_{b} \sqrt[4]{\frac{a}{b}}}) \sqrt{\log_{a} b}}$

Moderate

Question

The value of $E if b \geq a > 1$ is

A

1
B

2
C

$2^{\log_{a} b}$
D

$2^{\log_{b} a}$

Solution

We have
$\begin{matrix} E = 2^{(\sqrt{\log_{a} \sqrt[4]{ab} + \log_{b} \sqrt[4]{ab}} - \sqrt{\log_{a} \sqrt[4]{\frac{b}{a}} + \log_{b} \sqrt[4]{\frac{a}{b}}}) \sqrt{\log_{a} b}} \\ = 2^{\frac{1}{2} (\sqrt{\log_{a} ab + \log_{b} ab} - \sqrt{\log_{a} b / a + \log_{b} a / b}) \sqrt{\log_{a} b}} \\ = 2^{\frac{1}{2} (\sqrt{2 + \log_{a} b + \log_{b} a} - \sqrt{\log_{a} b + \log_{b} a - 2}) \sqrt{\log_{a} b}} \\ = 2^{\frac{1}{2} (\sqrt{{(\log_{a} b)}^{2} + 2 \log_{a} b + 1} - \sqrt{{(\log_{a} b)}^{2} - 2 \log_{a} b + 1})} \\ = 2^{\frac{1}{2} (\sqrt{{(\log_{a} b + 1)}^{2}} - \sqrt{{(\log_{a} b - 1)}^{2}})} \\ = 2^{\frac{1}{2} (|\log_{a} b + 1| - |\log_{a} b - 1|)} \end{matrix}$
Case I:
$\begin{array}{l} b \geq a > 1 \\ \Rightarrow \log_{a} b \geq \log_{a} a \\ \Rightarrow \log_{a} b \geq 1 \\ \Rightarrow E = 2^{\frac{1}{2} (\log_{a} b + 1 - \log_{a} b + 1)} = 2 \end{array}$
Case II:
$\begin{array}{l} 1 < b < a \\ \Rightarrow 0 < \log_{a} b < \log_{a} a \\ \Rightarrow 0 < \log_{a} b < 1 \\ \Rightarrow E = 2^{\frac{1}{2} (\log_{a} b + 1 - 1 + \log_{a} b)} \\ = 2^{1 / 2 \cdot (2 \log_{a} b)} \\ = 2^{\log_{a} b} \end{array}$

Question

The value of E if 1<b<a is

A

1
B

2
C

$2^{\log_{a} b}$
D

$2^{\log_{b} a}$

Solution

We have
$\begin{matrix} E = 2^{(\sqrt{\log_{a} \sqrt[4]{ab} + \log_{b} \sqrt[4]{ab}} - \sqrt{\log_{a} \sqrt[4]{\frac{b}{a}} + \log_{b} \sqrt[4]{\frac{a}{b}}}) \sqrt{\log_{a} b}} \\ = 2^{\frac{1}{2} (\sqrt{\log_{a} ab + \log_{b} ab} - \sqrt{\log_{a} b / a + \log_{b} a / b}) \sqrt{\log_{a} b}} \\ = 2^{\frac{1}{2} (\sqrt{2 + \log_{a} b + \log_{b} a} - \sqrt{\log_{a} b + \log_{b} a - 2}) \sqrt{\log_{a} b}} \\ = 2^{\frac{1}{2} (\sqrt{{(\log_{a} b)}^{2} + 2 \log_{a} b + 1} - \sqrt{{(\log_{a} b)}^{2} - 2 \log_{a} b + 1})} \\ = 2^{\frac{1}{2} (\sqrt{{(\log_{a} b + 1)}^{2}} - \sqrt{{(\log_{a} b - 1)}^{2}})} \\ = 2^{\frac{1}{2} (|\log_{a} b + 1| - |\log_{a} b - 1|)} \end{matrix}$
Case I:
$\begin{array}{l} b \geq a > 1 \\ \Rightarrow \log_{a} b \geq \log_{a} a \\ \Rightarrow \log_{a} b \geq 1 \\ \Rightarrow E = 2^{\frac{1}{2} (\log_{a} b + 1 - \log_{a} b + 1)} = 2 \end{array}$
Case II:
$\begin{array}{l} 1 < b < a \\ \Rightarrow 0 < \log_{a} b < \log_{a} a \\ \Rightarrow 0 < \log_{a} b < 1 \\ \Rightarrow E = 2^{\frac{1}{2} (\log_{a} b + 1 - 1 + \log_{a} b)} \\ = 2^{1 / 2 \cdot (2 \log_{a} b)} \\ = 2^{\log_{a} b} \end{array}$

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