Show that ${10^{\log x}} = x$?

Hint: Take the left-hand side of the expression and assume it equal to $y$. After that take the log on both sides. After that use the log property of power, $\log {a^b} = b\log a$. After that substitute the value of $\log 10$. Then, cancel out the log from both sides to arrive at the right-hand side of the expression.

Complete step-by-step answer:
Let us understand the definition of log first.
Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.
In other words, a logarithm is essentially an exponent that is written in a particular manner.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
There are two types of logarithms which is the common logarithmic function and the natural logarithmic function.
Common Logarithmic Function or Common logarithm is the logarithm with a base equal to 10. It is also known as the decimal logarithm because of its base. The common logarithm of x is denoted as \[\log x\]. For example, $\log 100 = 2$ (Since, ${10^2} = 100$).
The Natural Logarithmic Function is the logarithm with a base equal to the mathematical constant $e$. The value of \[e\] which is a mathematical constant is approximately equal to 2.7182818. The common logarithm of x is denoted as \[{\log _e}x\] or $\ln x$. For example, ${\log _e}100 = \ln 100$.
Let the left-hand side of the expression be $y$.
$ \Rightarrow y = {10^{\log x}}$
Take log on both sides,
$ \Rightarrow \log y = \log {10^{\log x}}$
We know that the power law of log is,
$\log {a^b} = a\log b$
Using the above law, the expression will be,
$ \Rightarrow \log y = \log x \times \log 10$
The value of ${\log _{10}}10$ is 1.
$ \Rightarrow \log y = \log x \times 1$
Simplify the terms,
$ \Rightarrow \log y = \log x$
Cancel out log on both sides,
$ \Rightarrow y = x$
The above result is the same as the right-hand side of the statement.

Hence it is proved.

Note:
The different rules of the log are,
Change of base rule law,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
Product rule law,
$\log xy = \log x + \log y$
Quotient rule law,
$\log \dfrac{x}{y} = \log x - \log y$
Power rule law,
$\log {x^y} = y\log x$