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If $7^{1 + x} + 7^{1 - x} = 50$ , then x is

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$1, - 1$

1,0

$- 1, - 1$

1,1

answer is C.

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Detailed Solution

$7^{1 + x} + 7^{1 - x} = 50$

$7^{1} {.7}^{x} + \frac{7^{1}}{7^{x}} = 50$

${7.7}^{2 x} + \frac{7^{1}}{7^{x}} = 50$

${7.7}^{2 x} + 7 - {50.7}^{x} = 0$

Let $7^{x} = a$

$\begin{array}{l} 7 a^{2} - 50 a + 7 = 0 \\ 7 a^{2} - 49 a - a + 7 = 0 \\ (7 a - 1) (a - 7) = 0 \end{array}$

$a = \frac{1}{7}$ or 7

So,

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