(2 log 6+6 log 2)/(4log2+log 27-log9)

What is the value this?

Good,

Answer = ?

(2Log6 +6log2)/(4log2+log27-log9)

2log6 = 2log(3×2)

2log6 = 2.(Log3 +Log2)

2Log6 = 2Log3+2Log2


Then,

2Log6+6log2 = 2Log3+2Log2+6log2

= 2Log3+8Log2

= 2(Log3+4Log2)


We have now:

Log27 = Log(3^3)

Log27 = 3Log3

And

Log9 = Log(3^2)

Log9 = 2Log3

Then us stay

4log2+log27-log9= 4log2+3log3-2log3

= 4log2 +log3

Soon,

Ans =2( Log3 +4Log2)/(4log2+log3)

We can to cut

Answer = 2