Pertanyaan

Nilai dari = ....

Nilai dari $begin mathsize 14px style open parentheses fraction numerator begin display style log presubscript space presuperscript 5 invisible function application space 9. log presubscript space presuperscript 81 invisible function application space 625 plus log presubscript space presuperscript 5 space invisible function application 125 end style over denominator begin display style log presubscript space presuperscript 6 space invisible function application 216 minus log presuperscript 6 space 36 end style end fraction close parentheses cubed end style$ = ....

625
125
25
-25
-125

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$begin mathsize 14px style open parentheses fraction numerator begin display style log presubscript space presuperscript 5 invisible function application space 9. log presubscript space presuperscript 81 invisible function application space 625 plus log presubscript space presuperscript 5 space invisible function application 125 end style over denominator begin display style log presubscript space presuperscript 6 space invisible function application 216 minus log presuperscript 6 space 36 end style end fraction close parentheses cubed end style$
$begin mathsize 14px style equals open parentheses fraction numerator log presubscript space presuperscript 5 invisible function application space 3 squared. log presubscript space presuperscript 3 to the power of 4 end presuperscript space invisible function application 5 to the power of 4 plus log presubscript space presuperscript 5 space invisible function application 5 cubed over denominator log presubscript space presuperscript 6 invisible function application space open parentheses 216 over 36 close parentheses end fraction close parentheses cubed equals open parentheses fraction numerator 2. begin display style 4 over 4 end style. log presubscript space presuperscript 5 space invisible function application 3. log presubscript space presuperscript 3 space invisible function application 5 plus 3. log presubscript space presuperscript 5 space invisible function application 5 over denominator log presubscript space presuperscript 6 space invisible function application 6 end fraction close parentheses cubed equals open parentheses fraction numerator 2. log presubscript space presuperscript 5 invisible function application space 5 plus 3 over denominator 1 end fraction close parentheses cubed equals open parentheses fraction numerator 2 plus 3 over denominator 1 end fraction close parentheses cubed equals 5 cubed equals 125 end style$