First slide

Trigonometric Identities

Question

If the equation $2^{x} + 4^{y} = 2^{y} + 4^{x}$ is solved for y in terms of x, is

Moderate

A

$x \log_{2} (1 - 2^{x})$
B

$x + \log_{2} (1 - 2^{x})$
C

$\log_{2} (1 - 2^{x})$
D

$x \log_{2} (2^{x} + 1)$

Solution

$2^{2 y} - 2^{y} + 2^{x} (1 - 2^{x}) = 0$
$Putting 2^{y} = t, we get$
$t^{2} - t + 2^{x} (1 - 2^{x}) = 0$ , $where t_{1} = 2^{y_{1}} and t_{2} = 2^{y_{2}}$
$\begin{array}{l} t_{1} t_{2} = 2^{x} (1 - 2^{x}) \\ 2^{y_{1} + y_{2}} = 2^{x} (1 - 2^{x}) \\ y_{1} + y_{2} = x + \log_{2} (1 - 2^{x}) \end{array}$

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