Question

Solve the equation $7^{x+1}+7^{1-x}=50$

Answer 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 $y=\frac{1}{7}$

$\Rightarrow$ Thus $7^x=7^1$

$x=1$;

$y=\frac{1}{7}$

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

$\Rightarrow x=-1$

$\Rightarrow x=\pm 1$