Kembali lagi bersama saya di blog tetamatika, tetamatika memberikan berbagai administrasi guru, materi ajar, media pembelajaran yang berkaitan dengan pelajaran matematika. Kali ini saya akan membahas tentang materi Logaritma beserta contohnya Kelas X Semester 1.
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Dari hubungan tersebut, diperoleh :
![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t463bOPQc1sUX1C9bp7sKquc3TUWdm4l1pYu9gHGcC-DdcYqN9XqV7q0FpZt6JAJIP_xBTfuCUMBOhASVgAhgJPA3YE7ZZ7bfBdI6kVerVbc7OfcKm9I1OEp-PnQkAAE6heuxI-Bp0vmicEf0Dd9HaX9nMHYU9LYexAGJ_H3Um_MjLEZMvs6V-iHprbr4NGvr7AWXWbuq9AF7jlA6DG8ZncEThdFG0il7F_ornWg2M5Nyj=s0-d)
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![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uNJfZ9g8NQhV4-kLM2cHaf3Hrziz5DgLxmfK3ODxH6RNtb4BUv-lpsO-LpZXo2k1LAzfPQxVyioIAEolUyD76NBniPGiuqgxsbSUyr6EHcC9gMLnXBPb3cuWi7k4aSV40gzg-c1WJwpNEVD79OFbZ_rtYfTf04RzDYh49LDR2-d2Tj749hskOyNr6HKavTPM3sHhqQJerWe074a7CXzBTBYP0Cwz7JWTXunGA7zQro0w-gbsitipOp2GDwVsui6hmIA9YsTunEo_cgInjctkbiW1esH4tBsA=s0-d)
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Sederhanakanlah !
a. 2log 4 + 2log 8
b. 3log (1/9) + 3log 81
c. 2log 2
+ 2log 4
Jawab :
a. 2log 4 + 2log 8 = 2log 4 . 8 = 2log 32 = 5
b. 3log (1/9) + 3log 81= 3log (1/9). 81 = 3log 9 = 2
c. 2log 2
+ 2log 4
= 2log 2
.4
= 2log 16 = 4
b. 42log 3
c. 93log 4
Jawab :
a. 55log 8 = 8
b. 42log 3 = 22.2log 3 = 22log 32 = 9
c. 93log 4 = 32.3log 4 = 33log 42 = 16
PENGERTIAN DAN SIFAT LOGARITMA
Logaritma adalah kebalikan dari suatu perpangkatan. Jika a dan b bilangan positif dengan
maka berlaku :
Dari hubungan tersebut, diperoleh :
Sifat-sifat logaritma
Lihat Juga : Materi Ajar Matematika SMA/SMK/MA Kurikulum 2013 Revisi
CONTOH SOAL SIFAT LOGARITMA
Sifat 1
Sederhanakanlah !
a. 2log 4 + 2log 8
b. 3log (1/9) + 3log 81
c. 2log 2
Jawab :
a. 2log 4 + 2log 8 = 2log 4 . 8 = 2log 32 = 5
b. 3log (1/9) + 3log 81= 3log (1/9). 81 = 3log 9 = 2
c. 2log 2
Sifat 2
Sederhanakanlah!
a. 2log 16 – 2 log 8
b. log 1.000 – log 100
c. 3log 18 – 3log 6
Jawab :
a. 2log 16 – 2 log 8 = 2log (16/8) = 2log 2 = 1
b. log 1.000 – log 100 = log (1000/100) = log 10 = 1
c. 3log 18 – 3log 6 = 3log (18/6) = 3log 3 = 1
a. 2log 16 – 2 log 8
b. log 1.000 – log 100
c. 3log 18 – 3log 6
Jawab :
a. 2log 16 – 2 log 8 = 2log (16/8) = 2log 2 = 1
b. log 1.000 – log 100 = log (1000/100) = log 10 = 1
c. 3log 18 – 3log 6 = 3log (18/6) = 3log 3 = 1
a. 55log 8
b. 42log 3
c. 93log 4
Jawab :
a. 55log 8 = 8
b. 42log 3 = 22.2log 3 = 22log 32 = 9
c. 93log 4 = 32.3log 4 = 33log 42 = 16
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