How do you solve \[{\log _{10}}x = - 1\] ?

Hint: Here in this question, we have to find the value of x. the function is a logarithmic function with the base 10. The logarithmic function can be written in the form of exponential form. Then by converting to exponential form we can determine the value of x.

Complete step-by-step answer:
The function is a logarithmic function. If the function involves the word log then it is a logarithmic function. In logarithmic function we have two kinds namely, common logarithmic function and natural logarithmic function. In the common logarithmic function, the base value is 10. In the natural logarithmic function, the base value is \[e\] . Here in the question, we have a common logarithmic function.
In general, the logarithmic function can be written in the form of exponential form i.e., given as \[{\log _a}x = b\] -----(1) can be written as \[{a^b} = x\] ----(2).
 Here the value of a is 10 and the value of b is -1. By substituting the values in the exponential form i.e., equation (2) we have
 \[{10^{ - 1}} = x\]
The exponent of the above number is negative, it can be written in the form of fraction.
Therefore, we have \[ \Rightarrow x = \dfrac{1}{{10}}\]
We can also the verify,
Substitute the value of x in the question
Therefore, we have
 \[{\log _{10}}x = - 1\]
 \[ \Rightarrow {\log _{10}}\left( {\dfrac{1}{{10}}} \right) = - 1\]
Use the property \[\log \left( {\dfrac{a}{b}} \right) = \log a - \log b\] ,
 \[ \Rightarrow {\log _{10}}1 - {\log _{10}}10 = - 1\]
We know the value \[{\log _{10}}1 = 0\] and \[ \Rightarrow {\log _{10}}10 = 1\]
Then by substituting we have
 \[ \Rightarrow 0 - 1 = - 1\]
 \[ \Rightarrow - 1 = - 1\]
Hence LHS is equal to RHS. Hence verified.
Therefore \[x = \dfrac{1}{{10}}\] .
So, the correct answer is “ \[x = \dfrac{1}{{10}}\] ”.

Note: The logarithmic function is inverse of exponential function. In these types of questions, we should know whether the function is a common logarithmic function or natural logarithmic function. Because these two functions are having different values and they differ from each other. Hence by using Clark’s table also we can find the value.