First slide

Trigonometric Identities

Question

If x = 9 is one of the solutions of $\log_{e} (x^{2} + 15 a^{2}) - \log_{e} (a - 2) = \log_{e} (\frac{8 ax}{a - 2})$ then

Moderate

A

$a = \frac{3}{5}$
B

a=3
C

x=15
D

x=2

Solution

$\log_{e} (x^{2} + 15 a^{2}) - \log_{e} (a - 2) = \log_{e} (\frac{8 ax}{a - 2})$
$Here a > 2, \frac{8 ax}{a - 2} > 0$
$∴ x > 0$
$Now \frac{x^{2} + 15 a^{2}}{a - 2} = \frac{8 ax}{a - 2}$
$\begin{array}{ll} or x^{2} - 8 ax + 15 a^{2} = 0 \\ \Rightarrow x = 3 a, 5 a \end{array}$
$now, x = 9$
$\begin{array}{ll} \Rightarrow a = 3, \frac{9}{5} \\ But a > 2 \\ ∴ a = 3 \end{array}$
For a=3, x=9, 15

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