Question

((81^x-2)/(27^x))=9

Im getting a bit confused with this one, thanks for your help :)

Answer 1

81, 27 and 9 are all powers of 3.
Let's write them in terms of 3.

81 = 3^4

Therefore we can write,
`81^(x-2)=3^(4(x-2))`

and since 27 = 3^3 we can write,
`27^x=3^(3x)`

and 9 is 3^2.

So our equation looks like this,
`(3^(4(x-2)))/(3^(3x))=3^2`

Applying our exponent rule, division to subtraction, gives us,
`3^(4(x-2)-3x)=3^2`

And from this point, since the BASES are the same,
we can simply equate the exponents:

4(x-2)-3x = 2

and then solve for x.