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# NEED HELP ASAP

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One ordered pair (a,b) satisfies the two equations $$ab^4 = 12$$ and a^5b^5=7776. What is the value of a in this ordered pair?

Sorry, one is in LaTex. NEED HELP PLEASE

Web2.0CalcUser123  Oct 13, 2018
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$$ab^4 = 12,~a^5b^5=7776 \\ \text{we want to get rid of }b \text{ by dividing a power of the first equation by a power of the 2nd}\\ \dfrac{(ab^4)^5}{(a^5b^5)^4}=\dfrac{(12)^5}{(7776)^4} \\ \dfrac{a^5b^{20}}{a^{20}b^{20}} = \dfrac{1}{a^{15}} = \dfrac{(2^23)^5}{(2^53^5)^4}=\dfrac{2^{10}3^5}{2^{20}3^{20}}=\dfrac{1}{2^{10}3^{15}}\\ a = (2^{10}3^{15})^{1/15} = 2^{2/3}3$$

Rom  Oct 13, 2018
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Can you write that in regular form, please? Like 2^6 or something like that for example. And how do you write 2/3 for the power?

Guest Oct 15, 2018
edited by Guest  Oct 15, 2018
#3
+93658
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$$2^{2/3}=(2^2)^{1/3} = \sqrt[3]{2^2}=\sqrt[3]{4}$$

does that help?

Melody  Oct 15, 2018