9^2x = 27^(3x-2)

To solve for x, we will use the exponent properties to find x.

First we will rewrite the bases as powers of prime numbers.

We know that:

9 = 3^2

27 = 3^3

Let us substitute.

==> 3^2)^2x = (3^3)^(3x-2)

Now we know that x^a^b = x^ab

==> 3^(2*2x) = 3^(3*(3x-2)

==> 3^(4x) = 3^(9x-6)

Now that the bases are equal, then the powers are equal too.

==> 4x = 9x -6

==> -5x = -6

**==> x = 6/5**

We have to solve 9^2x = 27^(3x-2)

9^2x = 27^(3x-2)

=> 3^2^2x = 3^3^(3x - 2)

=> 3^ 4x = 3^(9x - 6)

As the base is the same on both the sides we can equate the exponent.

4x = 9x - 6

=> 5x = 6

=> x = 6/5

Therefore **x = 6/5**

Q: 9^2x = 27^(3x-2). We have to find x.

We notice that both sides have different bases. We convert both side to a common base 3.

LHS = 9 ^2x = (3^2)^(2x) = 3^4x.

RHS = 27^(3x-4) = {3^3}(3x-2) = 3^(9x-6), as (a^m)^n) = a^(mn) by the law of exponents.

Therefore the give equation becomes:

3^4x = 3^(9x-6).

Since the bases are common , we equate the exponents:

4x = 9x - 6

4x-9x = -6.

-5x = -6.

-5x/-5 = -6/-5.

x= 6/5.

**So x= 6/5.**