How do you write as a single log for #log_460 - log_44 + log_4x #?

1 Answer
Dec 11, 2015

Answer:

I think you intended #log_4 60 - log_4 4 + log_4 x#.

If so, then using the basic properties of logs this simplifies to:

#log_4 15x#

Explanation:

In general, #log_a b + log_a c = log_a bc#

and #log_a b - log_a c = log_a (b/c)#

So:

#log_4 60 - log_4 4 + log_4 x#

#=log_4 (60/4) + log_4 x#

#=log_4 15 + log_4 x#

#=log_4 15x#