Pertanyaan

Hasil dari adalah...

Hasil dari $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator blank to the power of size 14px 3 size 14px log size 14px space size 14px 36 size 14px times to the power of size 14px 6 size 14px log size 14px space size 14px 81 size 14px plus to the power of size 14px 4 size 14px log size 14px space size 14px 32 over denominator blank to the power of begin mathsize 14px display style 1 over 9 end style end exponent size 14px log size 14px space size 14px 27 end fraction end cell end table$ adalah...

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jadi, hasil dari persamaan tersebut adalah

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator blank cubed log space 36 times to the power of 6 log space 81 plus to the power of 4 log space 32 over denominator blank to the power of begin display style 1 over 9 end style end exponent log space 27 end fraction end cell equals cell fraction numerator blank cubed log space 6 squared times to the power of 6 log space 3 to the power of 4 plus to the power of 2 squared end exponent log space 2 to the power of 5 over denominator blank to the power of begin display style 3 to the power of negative 2 end exponent end style end exponent log space 3 cubed end fraction end cell row blank equals cell fraction numerator 2 space cubed log space 6 times to the power of 6 log space 3 to the power of 4 plus begin display style 5 over 2 end style space squared log space 2 over denominator negative begin display style 3 over 2 end style space cubed log space 3 end fraction end cell row blank equals cell fraction numerator 2 space cubed log space 3 to the power of 4 plus begin display style 5 over 2 times 1 end style over denominator negative begin display style 3 over 2 end style times 1 end fraction end cell row blank equals cell fraction numerator 2 times 4 space cubed log space 3 plus begin display style 5 over 2 end style over denominator negative begin display style 3 over 2 end style end fraction end cell row blank equals cell fraction numerator 2 times 4 space plus begin display style 5 over 2 end style over denominator negative begin display style 3 over 2 end style end fraction end cell row blank equals cell fraction numerator 8 plus begin display style 5 over 2 end style over denominator negative begin display style 3 over 2 end style end fraction end cell row blank equals cell fraction numerator begin display style 21 over 2 end style over denominator negative begin display style 3 over 2 end style end fraction end cell row blank equals cell 21 over 2 cross times left parenthesis negative 2 over 3 right parenthesis end cell row blank equals cell negative 7 end cell end table end style$

jadi, hasil dari persamaan tersebut adalah negative 7