Integral value of `x` which satisfies the equation `=log_6 54+(log)_x 16=(log)_(sqrt(2))x-(log)_(36)(4/9)i s ddot`

Answer

4

Explanation

`log_(6) 54+log_(x)16=log_(sqrt2)x-log_(36). 4/9`
` or 1+2 log_(6) 3+log_(x) 16 = 2 log_(2) x- log_(6). 2/3`
` or 1+2 log_(6) 3+log_(x) 16=2 log_(2)x-log_(6) 2+log_(6) 3`
`or 1+ log_(x) 16=2 log_(2) x- (log_(6)2+log_(6)3)`
`or 1+ log_(x) 16 = 2 log_(2)x-1`
` or 4/(log_(2)x) = 2 log_(2) x-2`
Let ` log_(2) x=t`,we have (t-2)(t+1) = 0
` rArr t = 2 or t=- 1`
`rArr x = 4 or 1//2`