Question

(2/3) log27-1/2 log(1/9)- 3log 3​

Answer 1

Answer:

0

Step-by-step explanation:

before doing the sum we need to know the

some basic fomulas

logaⁿ=nloga

log(ab)= loga +logb

log(a/b)=loga-logb

log1=0

so

now

given (2/3)log27 -(1/2)log(1/9) -3log3

27=3³

so

(2/3)(3)log3-(1/2)(2)log(1/3)-3log3

on simllfy

2log3-log(1/3)-3log3

-log3 -log(1/3)

-[log3+ log(1/3)]

-[log3*1/3]

-[log1]

0