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Question Number 66760 by John Kaloki Musau last updated on 19/Aug/19

By writing your answer in the  form a^y  simplify  (3^(5x) ×5^(2x) ×3^(−x) ÷5^(−2x) )^(1/4)

$${By}\:{writing}\:{your}\:{answer}\:{in}\:{the} \\ $$$${form}\:{a}^{{y}} \:{simplify} \\ $$$$\left(\mathrm{3}^{\mathrm{5}{x}} ×\mathrm{5}^{\mathrm{2}{x}} ×\mathrm{3}^{−{x}} \boldsymbol{\div}\mathrm{5}^{−\mathrm{2}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$

Commented by Prithwish sen last updated on 19/Aug/19

15^x

$$\mathrm{15}^{\mathrm{x}} \\ $$

Commented by John Kaloki Musau last updated on 20/Aug/19

how did u do it?

$${how}\:{did}\:{u}\:{do}\:{it}? \\ $$

Answered by Kunal12588 last updated on 20/Aug/19

(((3^(5x) ×5^(2x) ×3^(−x) )/5^(−2x) ))^(1/4)   =(3^(5x−x) ×5^(2x+2x) )^(1/4)   =(3^(4x) ×5^(4x) )^(1/4)   =15^(4x×(1/4))   =15^x

$$\left(\frac{\mathrm{3}^{\mathrm{5}{x}} ×\mathrm{5}^{\mathrm{2}{x}} ×\mathrm{3}^{−{x}} }{\mathrm{5}^{−\mathrm{2}{x}} }\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$=\left(\mathrm{3}^{\mathrm{5}{x}−{x}} ×\mathrm{5}^{\mathrm{2}{x}+\mathrm{2}{x}} \right)^{\mathrm{1}/\mathrm{4}} \\ $$$$=\left(\mathrm{3}^{\mathrm{4}{x}} ×\mathrm{5}^{\mathrm{4}{x}} \right)^{\mathrm{1}/\mathrm{4}} \\ $$$$=\mathrm{15}^{\mathrm{4}{x}×\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\mathrm{15}^{{x}} \\ $$

Commented by John Kaloki Musau last updated on 21/Aug/19

I like it

$${I}\:{like}\:{it} \\ $$

Answered by Mr Jor last updated on 25/Aug/19

(3^(5x−x) ×5^(2x−(−2x)) )^(1/4)   (3^(4x) ×5^(4x) )^(1/4) =3^(4x×(1/4)) ×5^(4x×(1/4))   3^x ×5^x                     =15^x

$$\left(\mathrm{3}^{\mathrm{5}\boldsymbol{{x}}−\boldsymbol{{x}}} ×\mathrm{5}^{\mathrm{2}\boldsymbol{{x}}−\left(−\mathrm{2}\boldsymbol{{x}}\right)} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\left(\mathrm{3}^{\mathrm{4}\boldsymbol{{x}}} ×\mathrm{5}^{\mathrm{4}\boldsymbol{{x}}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} =\mathrm{3}^{\mathrm{4}\boldsymbol{{x}}×\frac{\mathrm{1}}{\mathrm{4}}} ×\mathrm{5}^{\mathrm{4}\boldsymbol{{x}}×\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\mathrm{3}^{\boldsymbol{{x}}} ×\mathrm{5}^{\boldsymbol{{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{15}^{\boldsymbol{{x}}} \\ $$

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