Question

$log _{4}(log _{3}^9 )+log _{2} (log _{3} ^{81} )+log0,8(log _{16} ^{32} )$

Obs são logs, nao frações,o resultado no gabarito tem -5/2

Answer 1

Propriedades que utilizarei:

$log_{b}(a^{n})=n\cdot log_{b}(a)\\\\log_{(b^{n})}(a)=\frac{1}{n}\cdot log_{b}(a)\\\\log_{b}(b)=1$

Transformando 0,8 em fração:

$0,8=\dfrac{8}{10}=\dfrac{4}{5}$
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Vamos trabalhar nos logaritmandos dos logaritmos "principais":

$log_{3}(9)=log_{3}(3^{2})=2\cdot log_{3}(3)=2\cdot1=2\\\\log_{3}(81)=log_{3}(3^{4})=4\cdot log_{3}(3)=4\cdot1=4\\\\log_{16}(32)=log_{(2^{4})}(2^{5})=\frac{5}{4}\cdot log_{2}(2)=\frac{5}{4}$
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$log_{4}(log_{3}[9])=log_{4}[2]=log_{(2^{2})}(2)=\frac{1}{2}\cdot log_{2}(2)=\frac{1}{2}\\\\log_{2}(log_{3}[81])=log_{2}(4)=log_{2}(2^{2})=2\cdot log_2(2)=2\\\\log_{0,8}(log_{16}[32])=log_{(\frac{4}{5})}(\frac{5}{4})=log_{(\frac{4}{5})}(\frac{4}{5})^{-1}=-1\cdot log_{(\frac{4}{5})}(\frac{4}{5})=-1$

Logo:

$log_{4}(log_{3}^{9})+log_{2}(log_{3}^{81})+log_{0,8}(log_{16}^{32})=\frac{1}{2}+2+(-1)\\\\log_{4}(log_{3}^{9})+log_{2}(log_{3}^{81})+log_{0,8}(log_{16}^{32})=\frac{1}{2}+1\\\\log_{4}(log_{3}^{9})+log_{2}(log_{3}^{81})+log_{0,8}(log_{16}^{32})=\frac{1+2}{2}\\\\\boxed{\boxed{log_{4}(log_{3}^{9})+log_{2}(log_{3}^{81})+log_{0,8}(log_{16}^{32})=\frac{3}{2}}}$

Answer 2

               
Log 9 = x ==> 3^x = 9 ==> 3^x = 3^2 ==> x = 2
       3
Log 2 =  ==> 4^y = 2 ^1 ==> (2^2)^y = 2^1 ==> 2y=1 ==> y = 1/2
      4
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Log81 = a ==> 3^a = 81 ==> 3^a = 3^4 ==> a = 4
       3
Log4 =b ==> 2^b = 4 ==> 2^b = 2^2 ==> b=2
     2
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Log32 =c  ==> 16^c = 32==> (2^4)^c = 2^5 ==>4c=5 ==>c=5/4
      16

log5/4 = d ==>0,8^d = 5/4==> 0,8^d = (4/5)^-1 ==>0,8^d=0,8^-1==>d= -1
     0,8
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y = 1/2  ;  b = 2 ; d = - 1
  1 + 2 - 1 ==>   1+ 4 - 2 ==> 3
  2                            2              2