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How does one solve ${\log _{10}}18 - {\log _{10}}3x = {\log _{10}}2$ ?

seo-qna
Last updated date: 25th Feb 2024
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IVSAT 2024
Answer
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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 .