Wednesday, October 27, 2021

Soal dan Pembahasan Logaritma Kelas X Sukino Peminatan LKS 5 Bag A no 12

12. Jika a > 1, maka penyelesaian dari persamaan [alog(2x + 1)][3log √a] = 1 adalah ...

a. 1

b. 2

c. 3

d. 4

e. 5


Jawaban : D

[alog(2x + 1)][3log √a] = 1

[alog(2x + 1)][3log a1/2] = 1

[alog(2x + 1)][1/2 3log a] = 1

1/2 . 3log a . alog(2x + 1) = 1

1/2 . 3log (2x + 1) = 1

3log (2x + 1) = 2

3log (2x + 1) = 3log 32

2x + 1 = 32

2x + 1 = 9

2x = 8

x = 4

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