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Question Number 187877 by nadovic last updated on 23/Feb/23
Simplify completely ((256^(−(7/(16))) × 128^(9/(28)) )/(512^((17)/(36)) × 64^(−((11)/(12))) ))
$$\mathrm{Simplify}\:\mathrm{completely} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{256}^{−\frac{\mathrm{7}}{\mathrm{16}}} \:×\:\mathrm{128}^{\frac{\mathrm{9}}{\mathrm{28}}} }{\mathrm{512}^{\frac{\mathrm{17}}{\mathrm{36}}} \:×\:\mathrm{64}^{−\frac{\mathrm{11}}{\mathrm{12}}} } \\ $$
Answered by Rasheed.Sindhi last updated on 23/Feb/23
((256^(−(7/(16))) × 128^(9/(28)) )/(512^((17)/(36)) × 64^(−((11)/(12))) )) =((256^((1/8)∙((−7)/2)) ×128^((1/7)∙(9/4)) )/(512^((1/9)∙((17)/4)) ×64^((1/6)∙((−11)/2)) )) =((2^(−7/2) ×2^(9/4) )/(2^(17/4) ×2^(−11/2) )) =(2^(−(7/2)+(9/4)) /2^(((17)/4)−((11)/2)) )=(2^(−(5/4)) /2^(−(5/4)) )=2^(−(5/4)+(5/4)) =2^0 =1
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{256}^{−\frac{\mathrm{7}}{\mathrm{16}}} \:×\:\mathrm{128}^{\frac{\mathrm{9}}{\mathrm{28}}} }{\mathrm{512}^{\frac{\mathrm{17}}{\mathrm{36}}} \:×\:\mathrm{64}^{−\frac{\mathrm{11}}{\mathrm{12}}} } \\ $$$$=\frac{\mathrm{256}^{\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{−\mathrm{7}}{\mathrm{2}}} ×\mathrm{128}^{\frac{\mathrm{1}}{\mathrm{7}}\centerdot\frac{\mathrm{9}}{\mathrm{4}}} }{\mathrm{512}^{\frac{\mathrm{1}}{\mathrm{9}}\centerdot\frac{\mathrm{17}}{\mathrm{4}}} ×\mathrm{64}^{\frac{\mathrm{1}}{\mathrm{6}}\centerdot\frac{−\mathrm{11}}{\mathrm{2}}} }\:\: \\ $$$$=\frac{\mathrm{2}^{−\mathrm{7}/\mathrm{2}} ×\mathrm{2}^{\mathrm{9}/\mathrm{4}} }{\mathrm{2}^{\mathrm{17}/\mathrm{4}} ×\mathrm{2}^{−\mathrm{11}/\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}^{−\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{4}}} }{\mathrm{2}^{\frac{\mathrm{17}}{\mathrm{4}}−\frac{\mathrm{11}}{\mathrm{2}}} }=\frac{\mathrm{2}^{−\frac{\mathrm{5}}{\mathrm{4}}} }{\mathrm{2}^{−\frac{\mathrm{5}}{\mathrm{4}}} }=\mathrm{2}^{−\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{4}}} =\mathrm{2}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by nadovic last updated on 23/Feb/23
Wow.. Thank you Sir
$${Wow}..\:{Thank}\:{you}\:{Sir} \\ $$
Answered by HeferH last updated on 23/Feb/23
(((2^8 )^(−(7/(16))) ∙(2^7 )^(9/(28)) )/((2^9 )^((17)/(36)) ∙(2^6 )^(−((11)/(12))) )) = ((2^(−(7/2)) ∙2^(9/4) )/(2^((17)/4) ∙2^(−((11)/2)) )) = (2^((9/4)−(7/2)) /2^(((17)/4)−((11)/2)) ) = (2^(−(5/4)) /2^(−(5/4)) ) = 1
$$\frac{\left(\mathrm{2}^{\mathrm{8}} \right)^{−\frac{\mathrm{7}}{\mathrm{16}}} \centerdot\left(\mathrm{2}^{\mathrm{7}} \right)^{\frac{\mathrm{9}}{\mathrm{28}}} }{\left(\mathrm{2}^{\mathrm{9}} \right)^{\frac{\mathrm{17}}{\mathrm{36}}} \centerdot\left(\mathrm{2}^{\mathrm{6}} \right)^{−\frac{\mathrm{11}}{\mathrm{12}}} }\:=\:\frac{\mathrm{2}^{−\frac{\mathrm{7}}{\mathrm{2}}} \centerdot\mathrm{2}^{\frac{\mathrm{9}}{\mathrm{4}}} }{\mathrm{2}^{\frac{\mathrm{17}}{\mathrm{4}}} \centerdot\mathrm{2}^{−\frac{\mathrm{11}}{\mathrm{2}}} }\:=\:\frac{\mathrm{2}^{\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{7}}{\mathrm{2}}} }{\mathrm{2}^{\frac{\mathrm{17}}{\mathrm{4}}−\frac{\mathrm{11}}{\mathrm{2}}} }\:=\:\frac{\mathrm{2}^{−\frac{\mathrm{5}}{\mathrm{4}}} }{\mathrm{2}^{−\frac{\mathrm{5}}{\mathrm{4}}} }\:=\:\mathrm{1} \\ $$

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