Question

Question #4549a

Answer 1

`x = 2`

Explanation:

`9^x-9=8(3^x)`

`=> (3^x)^2-8(3^x)-9=0`

Let `u = 3^x`

`u^2-8u - 9 = 0`

`=> (u-9)(u+1)=0`

`=> u-9 = 0 or u+1 = 0`

`=> u = 9 or u = -1`

`=> 3^x = 9 or 3^x = -1`

As `3^x in (0, oo)` for `x in RR`, we can dismiss the second possibility

`=> 3^x = 9`

`=> 3^x = 3^2`

`:. x = 2`