Please solve (Rational Exponents)?

Question

The questions are:
27^x/9^2x-1=3^x+4

27^x(9^2x-1)=3^x+4

and
27^x+1=(1/9)^2x-5

I'd even be happy if you could just solve one:) Thank you!

Answer 1

I suppose that the first is 27^x / 9^(2x-1) = 3^(x+4)

27 = 3³ and 9 = 3²

then 27^x = (3³)^x = 3^(3x)
because the rule is (a^b)^c = a^(bc)

and 9^(2x-1) = 3^(4x-2)

27^x / 9^(2x-1) = 3^(3x) / 3^(4x-2) = 3^(3x-4x+2)
=3^(-x+2) because the rule is a^b / a^c = a^(b-c)

then 3^(-x+2) = 3^(x+4)

-x + 2 = x + 4
-2x = 2
x = -1

For the 2nd :
27^x * 9^(2x-1) = 3^(3x) * 3^(4x-2) = 3^(3x+4x-2)
=3^(7x-2) because the rule is a^b * a^c = a^(b+c)

then 7x - 2 = x + 4 and x = 1

For the 3rd :

1/9 = 3^(-2)
then the equation is

3^(3x+3) = 3^(-4x+10)

3x + 3 = -4x + 10

x = 1