Question

Question #b71c3

Answer 1

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`

Answer 2

`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.`

Answer 3

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.