# solve for x: 9^(x-1)=27^(x+1)

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We have to solve the equation 9^(x-1)=27^(x+1) for x.

9^(x-1)=27^(x+1)

=> 9^x/ 9 = 27^x * 27

=> 9^x / 27^x = 27*9

=> 3^2x / 3^3x = 3^5

=> 3^(2x - 3x) = 3^5

As the base 3 is the same we can equate the exponential

=> -x = 5

=> x = -5

**The required value of x is -5.**

To solve for x: 9^(x-1) = 27^(x+1).

We know that 9 = 3^2 and 27 = 3^3.

So LHS = 9^(x-1) =( 3^2)^(x-1) = 3^(2x-2), as (a^m)^n = a^(mn) by exponent law.

RHS = 27^(x+1) = (3^3)^(x+1) = 3^(3x+3), as(a^m)^n = a^(mn) by exponent law.

Therefore the given equation is 3^(2x-2) = 3^(3x+3). Since both sides have the same base, we equate the exponents:

2x-2 = 3x+3.

-2-3 = 3x-2x = x.

So x = -5.

**Therefore x = -5 is the solution.**

The answer is x= -5.