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Berapa 2log15, dan 4log75 ??
2log3=a, maka log3/log2=a, log3=alog2, log2=log3/a
3log5=b, maka log5/log3=b, log5=blog3
A. 2log15 = log15/log2
=log(3.5)/log2
=(log3+log5)/log2
=log3+blog3/(log3/a)
=(log3(1+b))/(log3(1/a))
=(1+b)/1/a
=a(1+b)
=a+ab
B. 4log75 = log75/log4
=log(5.5.3)/log(2.2)
=log5+log5+log3/log2+log2
=(blog3+blog3+log3)/(log3/a)+(log3/a)
=(log3(b+b+1))/(log3(2/a))
=(2b+1)/(2/a)
=a(2b+1)/2
=(2ab+a)/2