Question

Solve for $x:3^{x+2}+3^{-x}=10$

• A. $(3^x+1)(3^{x+2}+1)$
• B. $(3^x-1)(3^{x+2}-1)$
• C. $(3^x-1)(3^x+1)$
• D. $(3^x+1)(3^{x+2}-2)$

Answer 1

Answer: (B)

$(3^x-1)(3^{x+2}-1)$

We have, $3^{x+2}+3^{-x}=10$

$\Rightarrow 3^x{.3}^2+(3^x{)}^{-1}=10$

Let $3^x=u$, we get

$u{.3}^2+u^{-1}=10$

$\Rightarrow 9u+\frac{1}{u}=10$

$\Rightarrow 9u^2+1=10u$

$\Rightarrow 9u^2-10u+1=0$

$\Rightarrow 9u^2-9u-u+1=0$

$\Rightarrow 9u(u-1)-1(u-1)=0$

$\Rightarrow (9u-1)(u-1)=0$

$\Rightarrow 9u-1=0$ or $u-1=0$

$\Rightarrow u=\frac{1}{9}$ or $u=1$

$\Rightarrow 3^x=\frac{1}{3^2}$ or $3^x=1$

$\Rightarrow 3^x=3^{-2}$ or $3^x=3^0$

On comparing , we get

$x=-2$ or $x=0$