Log(16x^2)+ 2 log (1/x)?

Question

a) log 32 

b) log16x 

c) log 16 

d) log 32x

Answer 1

log(16x^2) + 2log(1/x)

= log((4^2)(x^2)) + 2log(1/x)

= log(4x)^2 + 2log(1/x)                 [Since loga^m = mloga ]

= 2log4x + 2log(1/x)

= 2 ( log4x + log(1/x) )

= 2 ( log4x * 1/x)                         [Since loga + logb = log a*b ]

=2 log 4x/x

= 2log 4

= log 4^2                                    [Since mloga = loga^m ]

= log 16

The answer is C.

Answer 2

The logarithm expression is

Apply power property of logarithm : .

Apply product property of logarithm : .

The answer is C.