Options

A. `{:(a,b,c,d),(s,p,q,r):}`
B. `{:(a,b,c,d),(s,p,q,r):}`
C. `{:(a,b,c,d),(q,r,p,s):}`
D. `{:(a,b,c,d),(r,p,q,s):}`

Answer

A

Explanation

a. `A=log_(2) log_(2) log_(4) 256 + log_(sqrt2)(sqrt2)^(4)`
` = log_(2) log_(2) log_(4) 4^(4) + 4`
` = log_(2) log_(2) 4+4=1+4 = 5`
b. `log_(3) (5x-2)-2log_(3) sqrt(3x+1) = 1 -log_(3) 4`
` or log_(3) (5x-2)-log_(3)(3x+1)+log_(3) 4=1`
`or log_(3) (((5x-2)(4))/(3x+1))=1`
` or ((5x-2)(4))/(3x+1) = 3`
` or x = 1`
c. ` 7^(log_(7)(x^(2)-4x+5))=(x-1)`
`or x^(2) - 4x + 5 = x-1`
` or x^(2) - 5x + 6 = 0`
` or (x-2)(x-3) = 0`
`rArr x = 2 or x = 3`
Also we must have ` x^(2) - 4x+5 gt 0 and x - 1 gt 0`
` rArr x gt 1 ("as " x^(2) - 4x + 5 gt 0 ` is true for all real numbers)
d. ` xgt 0, 1/2log_(2) x -2((log_(2)x)/2)^(2)+1gt0`
` rArr log_(2) x - (log_(2)x)^(2) + 2 gt 0`
` rArr (log_(2)x)^(2) - log_(2) x - 2 lt 0`
Let ` log_(2) x = t," we have " t^(2) - t-2 lt 0`
` rArr (t-2) (t+1) lt 0`
` rArr -1 lt t lt 2`
` rArr - 1 lt log_(2) x lt 2`
` rArr 1/2 lt x lt 4`
Hence, the number of intergers is 3, i.e., `{1, 2, 3}`.