First slide

Trigonometric Identities

Question

If $x_{1} and x_{2}$ are the roots of the equation $e^{2} \cdot x^{\ln x} = x^{3}$ with $x_{1} > x_{2},$ then

Moderate

A

$x_{1} = 2 x_{2}$
B

$x_{1} = x_{2}^{2}$
C

$2 x_{1} = x_{2}^{2}$
D

$x_{1}^{2} = x_{2}^{3}$

Solution

$e^{2} \cdot x^{\ln x} = x^{3}$
Taking log on both sides, we get
$\ln (e^{2} \cdot x^{\ln x}) = \ln (x^{3})$
$\begin{array}{ll} \Rightarrow (\ln x)^{2} - 3 \ln x + 2 = 0 \\ \Rightarrow (\ln x - 2) (\ln x - 1) = 0 \end{array}$
$If \ln x = 2 \Rightarrow x = e^{2}$
$If \ln x = 1 \Rightarrow x = e$
$Since x_{1} > x_{2}, we get x_{1} = e^{2} and x_{2} = e$
$\Rightarrow x_{1} = x_{2}^{2}$

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