How do you solve ${5^{ - x}} = 250$?

Hint:: In order to determine the value of the above question , there is a variable in the exponent that means we need to solve this question through Logarithm .
First take logarithm on both sides of the equation. Use the logarithmic exponential rule$\log {m^n} = n\log m$ . Thereafter, apply Change Of Base formula. At last by using a logarithmic calculator determine the logarithmic value . You will get your desired answer.
Formula Used:
$
n\log m = \log {m^n} \\
{m^{\log (n)}} = n \\
$

Complete step by step solution:
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.
First, we are going to rewrite the number
${5^{ - x}} = 250$
Now taking logarithm on both the sides
$\log \left( {{5^{ - x}}} \right) = \log \left( {250} \right)$
Using proper of logarithm $\log {m^n} = n\log m$
So,
\[
- x\log \left( {{5^{}}} \right) = \log \left( {250} \right) \\
\\
\]
Dividing both sides by $\log (5)$
\[ \Rightarrow - x = \dfrac{{\log (250)}}{{\log (5)}}\]
By Applying the Change Of Base formula $\dfrac{{\log (a)}}{{\log (b)}} = {\log _b}(a)$
\[ \Rightarrow - x = {\log _5}(250)\]
Divide both sides by -1.
\[ \Rightarrow x = - {\log _5}(250)\]
Now using calculator calculating the value of ${\log _5}(250)$which comes to be
${\log _5}(250) = 3.430676558$
Therefore, the solution is$x = - 3.430676558$.
Change Of Base formula $\dfrac{{\log (a)}}{{\log (b)}} = {\log _b}(a)$
${\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)$
Additional Information:
Value of the constant ”e” is equal to 2.71828.
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.
Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$
Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
${\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)$
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: Don’t forget to cross check your result. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$
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't be disentangled on the grounds that the bases (the x and y)
are not the equivalent.