How do you solve $2{\log _7}( - 2r) = 0$?

Hint:In order to determine the value of the above question, we will convert the expression into exponential form and to do so use the definition of logarithm that the logarithm of the form ${\log _b}x = y$ is when converted into exponential form is equivalent to ${b^y} = x$, so compare with the given logarithm value with this you will get your required answer.

Complete step by step solution:
We are given a logarithmic equation $2{\log _7}( - 2r) = 0$.
Let’s divide the equation with the number $2$,we get
$
\dfrac{2}{2}{\log _7}( - 2r) = 0 \times \dfrac{1}{2} \\
{\log _7}( - 2r) = 0 \\
$
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
Lets convert this into its exponential form to remove ${\log _7}$.
Any logarithmic form ${\log _b}X = y$when converted into equivalent exponential form results in
${b^y} = X$
So in Our question we are given ${\log _e}(\ln x) = 1$and if compare this with ${\log _b}x = y$we get
$
b = 7 \\
y = 0 \\
X = - 2r \\
$
$
\Rightarrow {\log _7}( - 2r) = 0 \\
\Rightarrow - 2r = {7^0} \\
$
As we know ,any number raised to the power zero is equal to 1.So ${7^0} = 1$
$
\Rightarrow - 2r = 1 \\
\Rightarrow r = - \dfrac{1}{2} \\
$
Therefore, solution to equation $2{\log _7}( - 2r) = 0$ is equal to $r = - \dfrac{1}{2}$

Additional Information:
1. Value of the constant” e” is equal to 2.71828.
2. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any
number , we actually undo an exponentiation.
3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
${\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)$
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$
6. The above guidelines work just if the bases are the equivalent. For example, the expression ${\log _d}(m) + {\log _b}(n)$can't be improved, on the grounds that the bases (the "d" and the "b") are not
the equivalent, similarly as x2 × y3 can'to be disentangled on the grounds that the bases (the x and y) are not the equivalent.

Note:
1. Don’t forgot to cross check your result
2. $\ln $ is known as natural logarithm
3.Logarithm of constant 1 is equal to zero.