9^2x = 27^(3x-2) find x.
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calendarEducator since 2008
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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
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
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.
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