How do you write the logarithmic expression as the sum, difference, or multiple of logarithms and simplify as much as possible for #log_5(25/x)#?

1 Answer
Jul 5, 2015

Answer:

#log_5(25/x) = 2 - log_5x#

Explanation:

Property: #log_b(x/y) = log_b x – log_b y#

So

#log_5(25/x) = log_5(25) – log_5x#

#log_5(25/x) = log_5(5^2) – log_5x#

Property: #log_b(x^d) =dlog_b x#

So

#log_5(25/x) = log_5(5^2) – log_5x = 2log_5(5) – log_5x#

#log_5(25/x) = = (2×1) – log_5x#

#log_5(25/x) = = 2 – log_5x#