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Answer 1 · 2016-07-10T09:39:49

See explanation.

Starting equation is #2log_7(-2r)=0#

First we can divide both sides by #2#

#log_7(-2r)=0#

Now we have to write the right side (0) as a logarythm.
Since #7^0=1#, we can write #0# as #log_7 1#, so the equation becomes:

#log_7(-2r)=log_7 1#

We can skip #log# signs because both sides are logarythms and the base is the same:

#-2r=1#

Now if we divide the equation by (-2) we get the answer:

#r=-1/2#

Answer 2 · 2016-07-10T09:44:31

# r=-1/2.#

I preume that the problem is #: 2log_7^(-2r)=0.#

#:. log_7^(-2r)=0.#

Now, by defn. of #log# fun., this means that #7^0=-2r.#

#:. 1=-2r.#

#:. r=-1/2.#

Answer 3 · 2016-07-11T01:07:55

Given equation
#2log_7(-2r) =0#

We know that #log# of any #-ve# number is not defined, as such #r# must be #-ve#

On the LHS we have two factors. Equating each with #0#, we see that #2!=0#. Therefore,
#log_7(-2r) =0#

By definition of #log# we obtain
#7^0=(-2r)#
#=>1=-2r#
#=>r=-1/2#
You have your answer.