Question

3 log 2 = x dan 2 log 5= y
12 log 15 =
bantu ya

Answer 1

¹²log 15 = ²log15 / ²log12 = ²log(3×5) / ²log(2²×3)
= (²log3+²log5) / (²log2²+²log3)
= (²log3+²log5) / (2ײlog2+²log3)
= (1/³log2+²log5) / (2ײlog2+1/³log2)
= (1/x + y) / (2×1 + 1/x)
= (1/x + y) / (2 + 1/x)
= (1/x + xy/x) / (2x/x + 1/x)
= ((1 + xy)/x) / ((2x + 1)/x)
= (1-xy)/(2x+1)