Solve 3(3^x) + 9(3^-x) = 28?

Question

can someone show me the steps please, thanks.

Answer 1

I'd start by substituting a = 3^x.

For the middle part, remember that 3^-x = 1/(3^x) = 1/a
3a + 9/a = 28

Now multiply through by a, to get rid of the denominator:
3a² + 9 = 28a

Put everything on the left side:
3a² - 28a + 9 = 0

Factor:
(3a - 1)(a - 9)

So a = 9 or a = 1/3

Now plug a = 3^x back in and solve from there.

3^x = 9 or 3^x = 1/3

3^2 = 9 or 3^-1 = 1/3

So the answers are x = 2 or x = -1

As a double check, try the answers:
x = 2:
3(3²) + 9(3^-2) =? 28
3(9) + 9(1/3²) =? 28
27 + 9(1/9) =? 28
27 + 1 = 28 <-- check

x = -1:
3(3^-1) + 9(3^-(-1)) =? 28
3(1/3) + 9(3^1) =? 28
1 + 9(3) =? 28
1 + 27 = 28 <-- check

So your two answers are:
x = 2 or x = -1

Answer 2

3(3^x)+9/3^x=28
put 3^x=u
3u+9/u=28
multiplying by u
3u^2+9=28u
adding -28u
3u^2-28u+9=0
3u^2-27u-u+9=0
3u(u-9)-1(u-9)=0
(u-9)(3u-1)=0
u=9
3^x=9=3^2
so x=2
u=1
3^x=1=3^0
x=0
so the solutions are x= 0 or 2

Answer 3

I am not sure what that little mark (up arrow) is would you clarify so that I can come back and work the problem?

Wow the next answer is totally insane. I guess I didn't get that far in math, my bad. Sorry for taking the two points.

Answer 4

DUH!!!! the answere is 28!!! that was sooo easy you just look at it and you got it just do it on your calculater!!!!