# Question #b71c3

Jul 10, 2016

See explanation.

#### Explanation:

Starting equation is $2 {\log}_{7} \left(- 2 r\right) = 0$

First we can divide both sides by $2$

${\log}_{7} \left(- 2 r\right) = 0$

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:

${\log}_{7} \left(- 2 r\right) = {\log}_{7} 1$

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

$- 2 r = 1$

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

$r = - \frac{1}{2}$

Jul 10, 2016

$r = - \frac{1}{2.}$

#### Explanation:

I preume that the problem is $: 2 {\log}_{7}^{- 2 r} = 0.$

$\therefore {\log}_{7}^{- 2 r} = 0.$

Now, by defn. of $\log$ fun., this means that ${7}^{0} = - 2 r .$

$\therefore 1 = - 2 r .$

$\therefore r = - \frac{1}{2.}$

Jul 11, 2016

Given equation
$2 {\log}_{7} \left(- 2 r\right) = 0$

We know that $\log$ of any $- v e$ number is not defined, as such $r$ must be $- v e$

On the LHS we have two factors. Equating each with $0$, we see that $2 \ne 0$. Therefore,
${\log}_{7} \left(- 2 r\right) = 0$

By definition of $\log$ we obtain
${7}^{0} = \left(- 2 r\right)$
$\implies 1 = - 2 r$
$\implies r = - \frac{1}{2}$