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^{9 + 18z})(81

^{-18z})(27

^{15 - 9z}) = 1, what is the value of z?

**D**

- Rewrite the terms on the left-side with common prime bases.

(243^{9 + 18z})(81^{-18z})(27^{15 – 9z})

= (3^{5(9 + 18z)})(3^{4(-18z)})(3^{3(15 – 9z)})

= (3^{45 + 90z})(3^{-72z})(3^{(45 – 27z)}) - Simplify by combining terms with like bases.

(3^{45 + 90z})(3^{-72z})(3^{(45 – 27z)})

= 3^{45 + 45 + 90z – 72z – 27z}

= 3^{90 -9z} - As a result of this simplification, the equation is much easier to deal with.

3^{90 -9z}= 1 - At this point, the problem may appear unsolvable as different bases and exponents exist. However, remember that any number raised to the 0th power is 1. Consequently, 1 can be rewritten as 3
^{0}. The equation now reads:

3^{90 -9z}= 1

3^{90 -9z}= 1 = 3^{0} - Since the bases are the same, you can set the exponents equal to each other. The equation becomes:

90 – 9z = 0

90 = 9z

z = 10

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