Question

solve x: 3x+3-x=2
please help me​

Answer 1

Answer :

x = 0

Solution :

• Given : 3^x + 3^(-x) = 2
• To find : x = ?

We have ,

=> 3^x + 3^(-x) = 2

=> 3^x + 1/3^x = 2

=> (3^x)•(3^x) + 1 = 2•(3^x)

=> (3^x)² - 2(3^x) + 1 = 0 -------(1)

Putting 3^x = y , eq-(1) will reduce to ;

=> y² - 2y + 1 = 0

=> (y - 1)² = 0

=> y = 1

=> 3^x = 1

=> 3^x = 3^0

=> x = 0

Hence , x = 0

Answer 2

Answer:

It is given that,

3^x+3^-x=2

3^x+1/3^x=2

3^x²+1=2(3^x)

3^x²-2(3^x)+1=0

Let,y=3^x

y²-2y+1=0

y²-y-y+1=0

on solving ,y=1

So,3^x=y

3^x=1

3^x=3⁰

x=0

Step-by-step explanation:

Hope it helps you frnd........