Question #b71c3

3 Answers
Jul 10, 2016

Answer:

See explanation.

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#

Jul 10, 2016

Answer:

# r=-1/2.#

Explanation:

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.#

Jul 11, 2016

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.