# Solve for x the equation 3^x+9^x=2

hery | Student

3^x + 9^x = 2

=> 3^x(1+9^x) = 2

=> 3^x(1+9^x) = 1 * 2

=> 3^x = 1 and 1+9^x =2

=> 3^x = 1 ----> x = 0  and 1+ 9^x =2 ---> 9^x = 2-1

----> 9^x = 1

----> x = 0

Thus, the integer solution for this problem is x = 0

giorgiana1976 | Student

We'll create matching bases within equation, writting 9 as a power of 3:

3^2 = 9

We'll raise to x power both sides:

(3^2)^x = 9^x

We'll multiply the exponents from the left:

3^2x = 9^x

We'll re-write the equation:

3^x + 3^2x = 2

We'll substitute 3^x by t:

t + t^2 = 2

We'll re-arrange the terms and we'll subtract 2 both sides:

t^2 + t - 2 = 0

t1 = [-1+sqrt(1 + 8)]/2

t1 = (-1+3)/2

t1 = 1

t2 = (-1-3)/2

t2 = -4/2

t2 = -2

We'll put 3^x = t1 => 3^x = 1

We'll create matching bases both sides, writting 1 as 3^0:

3^x = 3^0

We'll use one to one property of exponentials:

x = 0

We'll put 3^x = t2 => 3^x = -2 impossible, since there is no real value of x such as 3^x = -2.

The unique solution of the given equation is x = 0.