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Question Number 5023 by love math last updated on 03/Apr/16

((25^(−1) + log_2 a^(−1) )/(5^(−1) −log_4 a))−(3/a^(−0.5) ) Simplify

$$\frac{\mathrm{25}^{−\mathrm{1}} +\:{log}_{\mathrm{2}} {a}^{−\mathrm{1}} }{\mathrm{5}^{−\mathrm{1}} −{log}_{\mathrm{4}} {a}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} } \\ $$$$ \\ $$$${Simplify} \\ $$

Answered by Rasheed Soomro last updated on 05/Apr/16

((25^(−1) + log_2 a^(−1) )/(5^(−1) −log_4 a))−(3/a^(−0.5) ) log_2 a^(−1) =((log_5 a^(−1) )/(log_5 2)) log_4 a=((log_5 a)/(log_5 4)) 25^(−1) =log_5 x⇒x=5^(25^(−1) ) 5^(−1) =log_5 x⇒x=5^5^(−1) ((25^(−1) + log_2 a^(−1) )/(5^(−1) −log_4 a))−(3/a^(−0.5) )=((log_5 (5^(25^(−1) ) )+((log_5 a^(−1) )/(log_5 2)))/(log_5 (5^5^(−1) )−((log_5 a)/(log_5 4))))−(3/a^(−0.5) ) =((log_5 (5^(25^(−1) ) )log_5 2+log_5 a^(−1) )/(log_5 2))×((log_5 4)/(log_5 (5^5^(−1) )log_5 4−log_5 a))−(3/a^(−0.5) ) ((log_5 (5^(25^(−log_5 2) ) )+log_5 a^(−1) )/(log_5 (5^5^(−log_5 ^ 4) )−log_5 a))×((log_5 4)/(log_5 2))−(3/a^(−0.5) ) ((log_5 (5^(25^(−log_5 2) ) ×a^(−1) ))/(log_5 (5^5^(−log_5 ^ 4) ÷log_5 a)))×((log_5 4)/(log_5 2))−(3/a^(−0.5) ) Too complicated,I can′t continue.

$$\frac{\mathrm{25}^{−\mathrm{1}} +\:{log}_{\mathrm{2}} {a}^{−\mathrm{1}} }{\mathrm{5}^{−\mathrm{1}} −{log}_{\mathrm{4}} {a}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} } \\ $$$${log}_{\mathrm{2}} {a}^{−\mathrm{1}} =\frac{{log}_{\mathrm{5}} {a}^{−\mathrm{1}} }{{log}_{\mathrm{5}} \mathrm{2}}\:\:\: \\ $$$${log}_{\mathrm{4}} {a}=\frac{{log}_{\mathrm{5}} {a}}{{log}_{\mathrm{5}} \mathrm{4}} \\ $$$$\mathrm{25}^{−\mathrm{1}} ={log}_{\mathrm{5}} {x}\Rightarrow{x}=\mathrm{5}^{\mathrm{25}^{−\mathrm{1}} } \\ $$$$\mathrm{5}^{−\mathrm{1}} ={log}_{\mathrm{5}} {x}\Rightarrow{x}=\mathrm{5}^{\mathrm{5}^{−\mathrm{1}} } \\ $$$$\frac{\mathrm{25}^{−\mathrm{1}} +\:{log}_{\mathrm{2}} {a}^{−\mathrm{1}} }{\mathrm{5}^{−\mathrm{1}} −{log}_{\mathrm{4}} {a}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} }=\frac{{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{25}^{−\mathrm{1}} } \right)+\frac{{log}_{\mathrm{5}} {a}^{−\mathrm{1}} }{{log}_{\mathrm{5}} \mathrm{2}}}{{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{5}^{−\mathrm{1}} } \right)−\frac{{log}_{\mathrm{5}} {a}}{{log}_{\mathrm{5}} \mathrm{4}}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} } \\ $$$$=\frac{{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{25}^{−\mathrm{1}} } \right){log}_{\mathrm{5}} \mathrm{2}+{log}_{\mathrm{5}} {a}^{−\mathrm{1}} }{{log}_{\mathrm{5}} \mathrm{2}}×\frac{{log}_{\mathrm{5}} \mathrm{4}}{{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{5}^{−\mathrm{1}} } \right){log}_{\mathrm{5}} \mathrm{4}−{log}_{\mathrm{5}} {a}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} } \\ $$$$\frac{{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{25}^{−{log}_{\mathrm{5}} \mathrm{2}} } \right)+{log}_{\mathrm{5}} {a}^{−\mathrm{1}} }{{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{5}^{−{log}_{\mathrm{5}} ^{} \mathrm{4}} } \right)−{log}_{\mathrm{5}} {a}}×\frac{{log}_{\mathrm{5}} \mathrm{4}}{{log}_{\mathrm{5}} \mathrm{2}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} } \\ $$$$\frac{{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{25}^{−{log}_{\mathrm{5}} \mathrm{2}} } ×{a}^{−\mathrm{1}} \right)}{{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{5}^{−{log}_{\mathrm{5}} ^{} \mathrm{4}} } \boldsymbol{\div}{log}_{\mathrm{5}} {a}\right)}×\frac{{log}_{\mathrm{5}} \mathrm{4}}{{log}_{\mathrm{5}} \mathrm{2}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} } \\ $$$$\mathrm{Too}\:\mathrm{complicated},\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{continue}. \\ $$

Answered by Rasheed Soomro last updated on 05/Apr/16

((25^(−1) + log_2 a^(−1) )/(5^(−1) −log_4 a))−(3/a^(−0.5) ) ((1−25log_2 a)/(25))×(5/(1−5log_4 a))−3(√a) ((1−25log_2 a)/(5−25log_4 a))−3(√a) ((1−25log_2 a)/(5−25(((log_2 a)/(log_2 4)))))−3(√a) ((1−25log_2 a)/(5−25(((log_2 a)/2))))−3(√a) ((1−25log_2 a)/(5−((25)/2)log_2 a))−3(√a)

$$\frac{\mathrm{25}^{−\mathrm{1}} +\:{log}_{\mathrm{2}} {a}^{−\mathrm{1}} }{\mathrm{5}^{−\mathrm{1}} −{log}_{\mathrm{4}} {a}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} } \\ $$$$\frac{\mathrm{1}−\mathrm{25}{log}_{\mathrm{2}} {a}}{\mathrm{25}}×\frac{\mathrm{5}}{\mathrm{1}−\mathrm{5}{log}_{\mathrm{4}} {a}}−\mathrm{3}\sqrt{{a}} \\ $$$$\frac{\mathrm{1}−\mathrm{25}{log}_{\mathrm{2}} {a}}{\mathrm{5}−\mathrm{25}{log}_{\mathrm{4}} {a}}−\mathrm{3}\sqrt{{a}} \\ $$$$\frac{\mathrm{1}−\mathrm{25}{log}_{\mathrm{2}} {a}}{\mathrm{5}−\mathrm{25}\left(\frac{{log}_{\mathrm{2}} {a}}{{log}_{\mathrm{2}} \mathrm{4}}\right)}−\mathrm{3}\sqrt{{a}} \\ $$$$\frac{\mathrm{1}−\mathrm{25}{log}_{\mathrm{2}} {a}}{\mathrm{5}−\mathrm{25}\left(\frac{{log}_{\mathrm{2}} {a}}{\mathrm{2}}\right)}−\mathrm{3}\sqrt{{a}} \\ $$$$\frac{\mathrm{1}−\mathrm{25}{log}_{\mathrm{2}} {a}}{\mathrm{5}−\frac{\mathrm{25}}{\mathrm{2}}{log}_{\mathrm{2}} {a}}−\mathrm{3}\sqrt{{a}} \\ $$$$ \\ $$

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