Question

2log3-3log2 how to solve

Answer 1

since alogb = log(b^a)

so using this property

2log3-3log2

=log9-log8

since loga-logb = log(a/b)

=log(9/8)

Answer 2

Answer:

Step-by-step explanation:

2log3=log(3^2)=log9

3log2=log(2^3)=log8

log9-log8=log(9/8)

log-logb=log(a/b)