Question

find (2 log6 +6log2)/(4log2+log27-log9) in esaay mathod

Answer 1

this is your required result result

Answer 2

Given : 2 log6 +6log2)/(4log2+log27-log9)

To Find : Value

Solution:

(2 log6 +6log2)/(4log2+log27-log9)

=(2 log(3x2) +6log2)/(4log2+log27-log9)

using log(ab) = log a + logb

log a - log b = log (a/b)

=( 2log3 + 2log2+ 6log2 )/(4log2 + log (27/9))

= (2log3 + 8log2)/(4log2 + log 3)

take 2 common in numerator

= 2(log 3 + 4log2) /(4log2 + log 3)

cancel log 3 + 4log2  from numerator and denominator

= 2

(2 log6 +6log2)/(4log2+log27-log9) = 2

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