(9^x +3^x - 2)^1/2 >= 9 - 3^x solve for x

Question

 

9^x - means 9 to the power x

>= means greater than and equal to

Answer 1

(9^x +3^x - 2)^1/2 ≥9 - 3^x

=>[(3^x)²+3^x-2]^1/2 ≥9 - 3^x................(1)

Let's assume 3^x=a and substituting in (1)

(a²+a-2)^1/2 ≥9-a

=>(a²+2a-a-2)^1/2 ≥9-a

=>[a(a+2)-(a+2)]^1/2≥9-a

=>[(a+2)(a-1)]^1/2≥9-a

=>(a+2)^1/2.(a-1)^1/2≥9-a

=> either, (a+2)^1/2≥9-a & (a-1)^1/2≤9-a -------------Case 1

or, (a+2)^1/2≤9-a & (a-1)^1/2≥9-a ---------------------Case 2

or, (a+2)^1/2≥9-a & (a-1)^1/2≥9-a --------------------Case 3

For a>9, 9-a<0, Case 1, Case 2 and Case 3 holds true.

For a=9, 9-a=0, none of Case 1, Case 2 and Case 3 holds true.

For a<9, 9-a>0, all of Case 1, Case 2 and Case 3 holds true.

Hence, either a>9 or a<9

thus 3^x>9 or 3^x<9

=> 3^x>3² or 3^x<3² 

=>x>2 or x<2 ... hence solution

 

Answer 2

√(9^x + 3^x - 2) ≥ 9 - 3^x

 

 

Square both sides of the equation to undo the square-root.

 

9^x + 3^x - 2 ≥ (9 - 3^x)²

 

9^x + 3^x - 2 ≥ (9 - 3^x)(9 - 3^x)

 

9^x + 3^x - 2 ≥ 81 - 18(3^x) + (3^x)²

 

(3²)^x + 3^x - 2 ≥ 81 - 18(3^x) + (3^x)²

 

(3)^2x + 3^x - 2 ≥ 81 - 18(3^x) + (3)^2x

 

 

Now we use substitution to make the inequality easier to solve.

 

Let       q = 3^x

 

           q² = (3)^2x

 

 

Using the substitution, we get

 

q² + q - 2 ≥ 81 - 18q + q²

 

 

Subtracting q² on both sides of the equation gives us

 

q - 2 ≥ -81 - 18q

 

 

Now we have a linear inequality, which is much easier to solve.

 

19q - 2 ≥ -81

 

19q ≥ -79

 

 

Now divide both sides of the inequality.

 

q ≥ -79/19

 

 

Now we substitute this value back into q=3^x

 

-79/19 = 3^x

 

 

Since 3 raised to any number is not a negative number, we pick a positive number that satisfies the solution of q.

 

q = 1

 

1 = 3^x

 

x = 0

 

 

Therefore the solution is        x≥0.

 

Now just verify this solution by plugging it into the original equation.