Jan 31, 2012 Thread starter #1 M MoneyKing New member Jan 31, 2012 4 simplify and answer should be in positive exponents. (((4x^6)^3(4y^-8))/((2x)^4(12y^3)^2))^1/2 please help and thanks
simplify and answer should be in positive exponents. (((4x^6)^3(4y^-8))/((2x)^4(12y^3)^2))^1/2 please help and thanks
Jan 31, 2012 #2 Also sprach Zarathustra Member Jan 31, 2012 54 Re: simplify MoneyKing said: simplify and answer should be in positive exponents. (((4x^6)^3(4y^-8))/((2x)^4(12y^3)^2))^1/2 please help and thanks Click to expand... $$ \huge{(}\frac{{(4x^6)^3}4y^{-8}}{(2x)^4(12y^3)^2}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}\frac{{(4^3x^{18})}4y^{-8}}{(2^4x^4)(12^2y^6)}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}\frac{{(64x^{18})}4y^{-8}}{(16x^4)(144y^6)}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}\frac{4x^{14}}{36y^{14}}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}\frac{x^{14}}{9y^{14}}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}(\frac{x}{9y})^{14}\huge{)}^{\frac{1}{2}} $$ $$ (\frac{x}{9y})^{7} $$
Re: simplify MoneyKing said: simplify and answer should be in positive exponents. (((4x^6)^3(4y^-8))/((2x)^4(12y^3)^2))^1/2 please help and thanks Click to expand... $$ \huge{(}\frac{{(4x^6)^3}4y^{-8}}{(2x)^4(12y^3)^2}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}\frac{{(4^3x^{18})}4y^{-8}}{(2^4x^4)(12^2y^6)}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}\frac{{(64x^{18})}4y^{-8}}{(16x^4)(144y^6)}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}\frac{4x^{14}}{36y^{14}}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}\frac{x^{14}}{9y^{14}}\huge{)}^{\frac{1}{2}} $$ $$ \huge{(}(\frac{x}{9y})^{14}\huge{)}^{\frac{1}{2}} $$ $$ (\frac{x}{9y})^{7} $$
Feb 1, 2012 #3 B battleman13 New member Jan 31, 2012 6 Re: simplify You should probably show any work you have tried first so that more importantly we can fix any misconceptions you may have about this process. If your going any further in math the ability to do the work in this problem will be required. You may now have the answer, but what you really need is the ability to reach it on your own.
Re: simplify You should probably show any work you have tried first so that more importantly we can fix any misconceptions you may have about this process. If your going any further in math the ability to do the work in this problem will be required. You may now have the answer, but what you really need is the ability to reach it on your own.
Feb 5, 2012 #4 S soroban Well-known member Feb 2, 2012 409 Hello, MoneyKing! $\text{Simplify: }\:\left[\dfrac{(4x^6)^3(4y^{-8})}{(2x)^4(12y^3)^2}\right]^{\frac{1}{2}}$ Click to expand... $\left[\dfrac{(4x^6)^3(4y^{-8})}{(2x)^4(12y^3)^2}\right]^{\frac{1}{2}} \;=\;\;\left[\dfrac{4^3(x^6)^3\cdot 4y^{-8}}{2^4x^4\cdot 12^2(y^3)^2}\right]^{\frac{1}{2}} \;=\;\;\left[\dfrac{64x^{18}\cdot 4y^{-8}}{16x^4\cdot144y^6}\right]^{\frac{1}{2}} $ . . . . . $=\;\;\left[\dfrac{x^{14}}{9y^{14}}\right]^{\frac{1}{2}} \;=\;\; \dfrac{(x^{14})^{\frac{1}{2}}}{9^{\frac{1}{2}}(y^{14})^{\frac{1}{2}}} \;=\;\;\dfrac{x^7}{3y^7} $
Hello, MoneyKing! $\text{Simplify: }\:\left[\dfrac{(4x^6)^3(4y^{-8})}{(2x)^4(12y^3)^2}\right]^{\frac{1}{2}}$ Click to expand... $\left[\dfrac{(4x^6)^3(4y^{-8})}{(2x)^4(12y^3)^2}\right]^{\frac{1}{2}} \;=\;\;\left[\dfrac{4^3(x^6)^3\cdot 4y^{-8}}{2^4x^4\cdot 12^2(y^3)^2}\right]^{\frac{1}{2}} \;=\;\;\left[\dfrac{64x^{18}\cdot 4y^{-8}}{16x^4\cdot144y^6}\right]^{\frac{1}{2}} $ . . . . . $=\;\;\left[\dfrac{x^{14}}{9y^{14}}\right]^{\frac{1}{2}} \;=\;\; \dfrac{(x^{14})^{\frac{1}{2}}}{9^{\frac{1}{2}}(y^{14})^{\frac{1}{2}}} \;=\;\;\dfrac{x^7}{3y^7} $