Pertanyaan

Nilai dari = ....

Nilai dari $begin mathsize 14px style fraction numerator log presuperscript 3 space 1 over 9 plus to the power of square root of 2 end exponent log space 9. space cubed log space 16 over denominator log presuperscript 2 space 10 minus squared log space 5 end fraction end style$ = ....

S. Intan

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Pembahasan

$begin mathsize 14px style fraction numerator log presuperscript 3 space 1 over 9 plus to the power of square root of 2 end exponent log space 9. space cubed log space 16 over denominator log presuperscript 2 space 10 minus squared log space 5 end fraction equals fraction numerator log presuperscript 3 space invisible function application 3 to the power of negative 2 end exponent plus log presuperscript 2 to the power of 1 half end exponent end presuperscript invisible function application space 3 squared. space log presuperscript 3 invisible function application space 2 to the power of 4 over denominator log presuperscript 2 space invisible function application 10 over 5 end fraction equals fraction numerator open parentheses negative 2 close parentheses space log presuperscript 3 invisible function application space 3 plus fraction numerator 2 over denominator 1 divided by 2 end fraction space log presuperscript 2 invisible function application space 3. open parentheses 4 space log presuperscript 3 invisible function application space 2 close parentheses space space over denominator log presuperscript 2 space invisible function application 2 end fraction equals fraction numerator open parentheses negative 2 close parentheses open parentheses 1 close parentheses plus 4 space log presuperscript 2 space invisible function application 3. open parentheses 4 space log presuperscript 3 invisible function application space 2 close parentheses over denominator left parenthesis 1 right parenthesis end fraction equals fraction numerator negative 2 plus 16 space log presuperscript 2 space invisible function application 2 over denominator open parentheses 1 close parentheses end fraction equals fraction numerator negative 2 plus 16 left parenthesis 1 right parenthesis over denominator left parenthesis 1 right parenthesis end fraction equals 14 end style$