SOLUTION: I need hep solving: (1/3)^x+1 = 9^x I made the bases the same: 9^-2(x+1) = 9^x Then I set the exponents as an equation: 2x + 2 = x 2 = -x X = -2 When I checked the ba

Question 563953: I need hep solving:
(1/3)^x+1 = 9^x
I made the bases the same:
9^-2(x+1) = 9^x
Then I set the exponents as an equation:
2x + 2 = x
2 = -x
X = -2
When I checked the back of my book for the answer I saw it was -1/3 and I don't how to get it.
Answer by TutorDelphia(193) (Show Source): You can put this solution on YOUR website!
(1/3)^x+1 = 9^x
you made a mistake making the bases the same:
9^-2=1/81 so that doesn't work
you want
9^-(1/2)=1/3
So you get:
9^-(1/2)(x+1) = 9^x
-(1/2)(x+1) simplifies to -1/2x-1/2 since you have to distribute the negative 1/2
-(1/2)x-1/2=x
Multiply both sides by 2 to clear the fraction
-x-1=2x
add x to both sides
-1=3x
divide both sides by 3
x=-1/3
Difficult problem!
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