Question

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

• A. $\pm 1$
• B. $\pm 3$
• C. $\pm 2$
• D. None

Answer 1

Answer: (A)

$\pm 1$

Given equation is: $7^{1+x}+7^{1-x}=50$

$\therefore {7.7}^x+7\times \frac{1}{7^x}=50$
Let, $7^x=y$
$\Rightarrow 7y+\frac{7}{y}=50\Rightarrow {7y}^2+7=50y$
$\Rightarrow {7y}^2-50y+7=0$
$\Rightarrow {7y}^2-49y-y+7=0$
$\Rightarrow (y-7)(7y-1)=0$

$\Rightarrow y=7$ and $y=\frac{1}{7}$
Thus $7^x=7^1$

$x=1$

$\Rightarrow 7^x=7^{-1}$

$\Rightarrow x=-1$
$\Rightarrow x=\pm 1$