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**Hint:**For simplifying the original equation , firstly used logarithm property $\log a - \log b = \log \dfrac{a}{b}$ then take base ten exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the equation.

**Formula used:**

We used logarithm properties i.e.,

$\log a - \log b = \log \dfrac{a}{b}$

And

${b^{{{\log }_b}a}} = a$ , the logarithm function says $\log x$ is only defined when $x$is greater than zero.

**Complete solution step by step:**

It is given that ,

${\log _{10}}18 - {\log _{10}}3x = {\log _{10}}2$ ,

We have to solve for $x$ .

Now using logarithm property $\log a - \log b = \log \dfrac{a}{b}$ ,

We will get,

${\log _{10}}\dfrac{{18}}{{3x}} = {\log _{10}}2$

Now , simplify the equation , we will get the following result ,

${\log _{10}}\dfrac{6}{x} = {\log _{10}}2$

Now , by assuming the base of the logarithm to be $10$ ,then take the base $10$ exponential of both sides of the equation, we will get the following result ,

For equation one ,

${10^{{{\log }_{10}}\left( {\dfrac{6}{x}} \right)}} = {10^{{{\log }_{10}}2}}$

By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get the following result ,

$\dfrac{6}{x} = 2$

Or

$x = 3$

Now recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero.

Therefore, in our original equation ${\log _{10}}18 - {\log _{10}}3x = {\log _{10}}2$ ,

Here,

$(3x) > 0$ ,

For $x = 3$ ,

$9 > 0$

**Therefore, we have our solutions i.e., $3$ .**

**Note:**The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero. While performing logarithm properties we have to remember certain conditions , our end result must satisfy the domain of that logarithm .

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