Question

solve for 3^x+2 + 3^-x= 10

Answer 1

Answer is in the attachment...x = 0 or x = -2

Answer 2

Given,

$3^{x} +2+3^{-x} =10$

To find,

Solving for x

Solution,

⇒ $3^{x+2} +\frac{1}{3^{x} } = 10$

⇒$3^{x}.3^{2}+\frac{1}{3^{x} } = 10$                    

Since ,     $x^{a+b}= x^{a} x^{b}$

⇒ Let $3^{x}$ be a.

⇒$9a+\frac{1}{a} = 10$

⇒$\frac{9a^{2}+1 }{a} = 10$

⇒ $9a^{2}+1=10a$

⇒ $9a^{2}-10a+1=0$

⇒ $9a^{2}-9a-1a+1=0$

⇒$9a(a-1)-(a-1)=0$

⇒$(a-1)(9a-1)=0$

⇒$a-1=0 \\a = 1$

⇒ $9a-1=0\\a=\frac{1}{9}$

Since a = $3^{x}$

Now equate a values ,a =1

$3^{x} = 1\\3^{x} = 3^{0} \\x = 0$

Next  $a = \frac{1}{9}$

$3^{x} = \frac{1}{9} \\\\3^{x} = \frac{1}{3} ^{2} \\3^{x} = 3^{-2}$

Since bases are same equate powers , x = -2.

Hence , the values of x are -2 and 0.