How do I solve #log_2 7 = log_2 (7/16)#?

I'm not sure if what I'm doing is correct, but I tried expanding it:

#log_2 7 - log_2 7 - log_2 16#
#= log_2 (7/7) - log_2 16#
#=log_2 1 - log_2 16#
#= log_2 (1/16)#
#=log_2 (1/2^4)#

and after that I just got stuck. Can someone help?

1 Answer
1s2s2p
Oct 15, 2017

The equation is the same as 4.

Explanation:

I'm guessing that based on your description that it is meant to be #log_2(7)-log_2(7/16)# and not #log_2(7)=log_2(7/16)#

There are two ways of doing this.

The way you did it is almost correct except you forgot to multiply by -1 after expanding #-log_2(7/16)#.

Remember, you are taking the negative of #log_2(7/16)#. The positive of #log_2(7/16)# is #log_2(7)-log_2(16)#, while the negative is #-(log_2(7)-log_2(16))=-log_2(7)+log_2(16)#

#log_2(7)-log_2(7/16)#
#log_2(7)-(log_2(7)-log_2(16))#
#log_2(7)-log_2(7)+log_2(16)#
#log_2(7/7)+log_2(16)#
#log_2(1)+log_2(16)#
#0+log_2(16)#
#=log_2(16)=4#

You know that #log_ab-log_ac=log_a(b/c)#

Well, #log_2(7)-log_2(7/16)#
#=log_2(7/(7/16))#
#=log_2((cancel(7)*16)/cancel(7))#
#=log_2(16)=4#

Related questions
See all questions in Common Logs
Impact of this question
2117 views around the world
You can reuse this answer
Creative Commons License