I'm trying to solve this problem step by step with positive exponents. I still can't seem to find the correct answer but here is the problem
[(2x^2)(z^-5))/(3z^4)]^-5
Divided By
[((3x^4)(z^9))/(2x^-6)]^3
You mean simplify this?
$\displaystyle \left(\frac{2x^2 z^{-5}}{3z^4}\right)^{-5} \div \left(\frac{3x^4 z^9}{2x^{-6}}\right)^{3}$
I'd take the reciprocal of the first fraction and change the outer exponent to a positive one:
$\displaystyle \left(\frac{3z^4}{2x^2 z^{-5}}\right)^5 \div \left(\frac{3x^4 z^9}{2x^{-6}}\right)^{3}$
Then I'd multiply the reciprocal of the second fraction (note that the outermost exponent does not change):
$\displaystyle \left(\frac{3z^4}{2x^2 z^{-5}}\right)^5 \times \left(\frac{2x^{-6}}{3x^4 z^9}\right)^3$
Move the remaining negative exponents:
$\displaystyle \left(\frac{3z^4 z^5}{2x^2}\right)^5 \times \left(\frac{2}{3x^4 z^9 x^6}\right)^3$
$\displaystyle \left(\frac{3z^9}{2x^2}\right)^5 \times \left(\frac{2}{3x^{10} z^9}\right)^3$
Apply the Power-of-a-Power property:
$\displaystyle \frac{3^5 z^{45}}{2^5 x^{10}} \times \frac{2^3}{3^3 x^{30} z^{27}}$
Multiply and simplify:
$\displaystyle \frac{2^3 3^5 z^{45}}{2^5 3^3 x^{40} z^{27}}$
$\displaystyle \frac{3^2 z^{18}}{2^2 x^{40}}$
$\displaystyle \frac{9z^{18}}{4x^{40}}$
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