How does one solve $\log 2 + \log x = 1$ ?

Hint:For simplifying the original equation , firstly use logarithm property $\log a + \log b = \log (ab)$ then take base $10$ exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ ,then simplifying the further and lastly recall that the logarithm function say ${\log _b}a$ is only defined when $a$ is greater than zero or a real positive number.

Formula used:
We used logarithm properties that are as follow,
First $\log a + \log b = \log (ab)$
And second ${b^{{{\log }_b}a}} = a$
We also use that for a logarithm function say ${\log _b}a$ is only defined when $a$ is greater than zero or a real positive number.

Complete step by step answer:
We have the following expression,
$\log 2 + \log x = 1$ ,
Use logarithm property that is $\log a + \log b = \log (ab)$ to simplify the expression ,we get,
$ \Rightarrow \log (2x) = 1$
By assuming the base of the logarithm to be $10$ ,then take the base $10$ exponential of both sides of the equation, we get,
$ \Rightarrow {10^{\log 2x}} = {10^1}$
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the expression, we get ,
$\Rightarrow 2x = 1 \\
\Rightarrow x = \dfrac{1}{2} \\ $
Now recall that the logarithm function says ${\log _b}a$ is only defined when $a$ is greater than zero or a real positive number. Therefore, in our original equation that is $\log 2 + \log x = 1$ , Here $x$ must be greater than zero or a real positive number.
$ \therefore x = \dfrac{1}{2} > 0$ ,

Therefore, we have our solution that is $\dfrac{1}{2}$.

Note:The logarithm function says $\log x$ is only defined when $x$ is greater than zero. Here we use the relationship between exponents and logarithm with this relationship so we can simplify many more problems. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one .