Pembahasan soal logaritma (4)


soal 1.
Jika {}^{5}\log 2 = p, nyatakan \log \frac{5}{2} dalam p.

Jawab.

jika {}^{5}\log 2 = p, maka dapat kita nyatakan dalam bentuk \frac{\log 2}{\log 5} = p

\frac{\log 2}{\log 5} + 1 = p + 1

\frac{\log 2}{\log 5}+\frac{\log 5}{\log 5} = p + 1

\frac{\log 2+\log 5}{\log 5} = p + 1

\frac{\log 10}{\log 5} = p + 1

\frac{1}{\log 5} = p + 1

\frac{1}{p+1} =\log 5 ……. (1)

kita ulang langkah di atas
\frac{\log 2}{\log 5} = p,

\frac{\log 5}{\log 2} = \frac{1}{p}

\frac{\log 5}{\log 2}+ 1 = \frac{1}{p} + 1

\frac{\log 5}{\log 2}+ \frac{\log 2}{\log 2} = \frac{1}{p} + \frac{p}{p}

\frac{\log 5+\log 2}{\log 2} = \frac{1+p}{p}

\frac{\log 10}{\log 2} = \frac{1+p}{p}

\frac{1}{\log 2} = \frac{1+p}{p}

\log 2 = \frac{p}{1+p} ….. (2)
Sehingga \log \frac{5}{2} =\log 5 - \log 2
subtitusi persamaan (1) dan (2) pada persamaan di atas

\log \frac{5}{2} = \frac{1}{p+1} - \frac{p}{1+p}

\log \frac{5}{2} = \frac{1-p}{1+p}

OK.

One thought on “Pembahasan soal logaritma (4)

  1. Bikin soalNya yAnk Banyak Agy dUnkzZZ,…….

    cLo aGy bYbutuHin ajj sEdIkit Ada’a giLiRan Nda dyButuhn ajj bUanyakkkk,……….

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