caranya dan jawabannya gimana?buat hari ini

3log 2 = m, 2log 7 = n
14log 54 = log 54/log 14
= log 3 x 3 x 3 x 2/ log 7x2
= 2log 3+2log3+2log3+2log2/2log7+2log2
=(1/m)+(1/m)+(1/m)+1/n+1
=3/m + 1/n+1
=(3+m/m)/n+1
=3+m/m(n+1)
=m+3/m(n+1)

jawabannya yg B

3log2 = m dan 2log7 = n
14log54 = (2log54)/(2log14)
= (2log27.2) / (2log7.2)
= (2log27 + 2log2) / (2log7 + 2log2)
= (2log3^3 + 1) / (n + 1)
= (3 . 2log3 + 1) / (n + 1)
= (3 . 1/m + 1) / (n + 1) . m/m
= (3 + m) / m(n + 1)
= (m + 3) / m(n + 1)