Matematika

Pertanyaan

4log28 = p, maka 7log2= ....
a. 2/p-1
b. 1/2(p-1)
c. p-1/ 2
d. 2p-2
e. p-1

1 Jawaban

  • [tex] ^4log28 = p \\ ^4log(4.7) = p \\ ^4log4 + ^4log7 = p \\ 1 + ^4log7 = p \\ ^4log7 = p - 1 \\ ^7log4 = \frac{1}{p - 1} \\ ^7log(2^2) = \frac{1}{p - 1} \\ 2.^7log2 = \frac{1}{p - 1} \\ ^7log2 = \frac{1}{2(p - 1)} [/tex]

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