Question

I need one more last help please. Thanks!

Answer 1

Hi Gary :)
I hope you are well today.

One rule of logarithms is
`log_(b) ( (m)/(n) ) = log_b(m)-log_b(n)`

The square root can be written as n to the 1/2 power, then
`√(n) = n^{ (1)/(2) }`

So rewrite
`log_b[( (57)/(74))^ (1)/(2) ]`

Following the rule we finally rewrite the logarithm as:


`(1)/(2) log_b(57)- (1)/(2) log_b(74)`

Thus, our answer is B