Ticky expression

**sgonzalez90** · June 2nd 2009, 08:46 PM

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

**yeongil** · June 2nd 2009, 09:38 PM

Originally Posted by sgonzalez90

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?
$\left(\frac{2x^2 z^{-5}}{3z^4}\right)^{-5} \div \left(\frac{3x^4 z^9}{2x^{-6}}\right)^{3}$

I'd take the reciprocal of the first fraction and change the outer exponent to a positive one:
$\left(\frac{3z^4}{2x^2 z^{-5}}\right)^5 \div \left(\frac{3x^4 z^9}{2x^{-6}}\right)^{3}$

Then I'd multiply the reciprocal of the second fraction (note that the outermost exponent does not change):
$\left(\frac{3z^4}{2x^2 z^{-5}}\right)^5 \times \left(\frac{2x^{-6}}{3x^4 z^9}\right)^3$

Move the remaining negative exponents:
$\left(\frac{3z^4 z^5}{2x^2}\right)^5 \times \left(\frac{2}{3x^4 z^9 x^6}\right)^3$

$\left(\frac{3z^9}{2x^2}\right)^5 \times \left(\frac{2}{3x^{10} z^9}\right)^3$

Apply the Power-of-a-Power property:
$\frac{3^5 z^{45}}{2^5 x^{10}} \times \frac{2^3}{3^3 x^{30} z^{27}}$

Multiply and simplify:
$\frac{2^3 3^5 z^{45}}{2^5 3^3 x^{40} z^{27}}$

$\frac{3^2 z^{18}}{2^2 x^{40}}$

$\frac{9z^{18}}{4x^{40}}$

01

**Amer** · June 2nd 2009, 09:43 PM

I am always late

**Amer** · June 2nd 2009, 09:45 PM

Thanks yeongil

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