Question

2log3=a, 3log10=b. Nilai 6log120?? mohon dibantu dgn caranya:-)

Answer 1

6log120= 6log6 + 6log20
           = 1 + 6log2 + 6log10
           = 1 + 3log2/3log6 + 3log10/3log6
           = 1 + 1/a/3log6 + b/3log6
           = 1 + 1/a/3log3+3log2 +b/3log3+3log2
           = 1 +1/a/1+1/a + b/1+1/a
           = 1 + a/a2+1 + b/a
           = 2 + a/a2 + b/a

Answer 2

2log3=y
3log5=x
3log10=log5.2/log3
3log5+3log2/3log3
=x+3log2 / 1
= x + 2 log 2/ 2 log3 / 1
x + 1/y