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Solve the equations ${\log _{100}}\left| {x + y} \right| = \dfrac{1}{2},$ ${\log _{10}}y - {\log _{10}}\left| x \right| = {\log _{100}}4$ for $x$ and $y$.

Last updated date: 09th Aug 2024
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Hint: This problem deals with logarithms. This problem is rather very easy and very simple, though it seems to be complex. In mathematics logarithms are an inverse function of exponentiation. Which means that the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x.
If given ${\log _b}x = a$, then $x$ is given by :
$\Rightarrow x = {b^a}$
In order to solve this problem basic formulas of logarithms are used here, such as:
$\Rightarrow \log a - \log b = \log \left( {\dfrac{a}{b}} \right)$
$\Rightarrow {\log _{{a^n}}}{b^m} = \dfrac{m}{n}{\log _a}b$

Given that ${\log _{100}}\left| {x + y} \right| = \dfrac{1}{2}$
Also given that ${\log _{10}}y - {\log _{10}}\left| x \right| = {\log _{100}}4$
Consider the equation ${\log _{100}}\left| {x + y} \right| = \dfrac{1}{2}$, as given below:
From the basic formula of logarithms, ${\log _{100}}\left| {x + y} \right| = \dfrac{1}{2}$, is expressed as given below:
$\Rightarrow {\log _{100}}\left| {x + y} \right| = \dfrac{1}{2}$
$\Rightarrow \left| {x + y} \right| = {\left( {100} \right)^{\dfrac{1}{2}}} = \sqrt {100}$
$\Rightarrow \left| {x + y} \right| = 10$
Now consider the equation ${\log _{10}}y - {\log _{10}}\left| x \right| = {\log _{100}}4$, as given below:
We know that ${\log _{10}}a - {\log _{10}}b = {\log _{10}}\left( {\dfrac{a}{b}} \right)$, applying it to the above equation as given below:
$\Rightarrow {\log _{10}}y - {\log _{10}}\left| x \right| = {\log _{100}}4$
$\Rightarrow {\log _{10}}\left( {\dfrac{y}{{\left| x \right|}}} \right) = {\log _{100}}4$
Also applying the another basic logarithmic formula which is ${\log _{{a^n}}}{b^m} = \dfrac{m}{n}{\log _a}b$, to the right hand side of the considered above equation, as given below:
$\Rightarrow {\log _{10}}\left( {\dfrac{y}{{\left| x \right|}}} \right) = {\log _{{{10}^2}}}{2^2}$
$\Rightarrow {\log _{10}}\left( {\dfrac{y}{{\left| x \right|}}} \right) = \dfrac{2}{2}{\log _{10}}2$
$\Rightarrow {\log _{10}}\left( {\dfrac{y}{{\left| x \right|}}} \right) = {\log _{10}}2$
On comparing the above equation on both sides, as given below:
$\Rightarrow \left( {\dfrac{y}{{\left| x \right|}}} \right) = 2$
$\Rightarrow y = 2\left| x \right|$
Thus we have two equations and two variables, which are $\left| {x + y} \right| = 10$ and $y = 2\left| x \right|$.
Now substitute the obtained expression of $y$, which is $y = 2\left| x \right|$ in the equation $\left| {x + y} \right| = 10$, as given below:
$\Rightarrow \left| {x + y} \right| = 10$
$\Rightarrow \left| {x + \left| {2x} \right|} \right| = 10$
Here two cases arise, where in the first case $x$ is greater than zero, which means $x$is positive.
In the second case $x$ is less than zero, which means that $x$ is negative.
Consider the first case where $x > 0$, as given below:
i) $x > 0$
Then $\left| {2x} \right|$ becomes $2x$, as $x$ is positive.
$\Rightarrow \left| {x + 2x} \right| = 10$
$\Rightarrow \left| {3x} \right| = 10$
$\Rightarrow 3\left| x \right| = 10$
As $x$ is positive, so $\left| {3x} \right|$ becomes $3x$, as given below:
$\Rightarrow 3x = 10$
$\Rightarrow x = \dfrac{{10}}{3}$
Now substitute the value of $x$ in the equation of $y$, which is $y = 2\left| x \right|$, as given below:
$\Rightarrow y = 2\left| x \right|$
As $x$ is positive, so $\left| {2x} \right|$ becomes $2x$, as given below:
$\Rightarrow y = 2x$
$\Rightarrow y = 2\left( {\dfrac{{10}}{3}} \right)$
$\therefore y = \dfrac{{20}}{3}$
Hence the values of $x$ and $y$ are $\left( {x,y} \right) = \left( {\dfrac{{10}}{3},\dfrac{{20}}{3}} \right)$
Consider the second case where $x < 0$, as given below:
ii) $x < 0$
Then $\left| {2x} \right|$ becomes $- 2x$, as $x$ is negative.
$\Rightarrow \left| {x - 2x} \right| = 10$
$\Rightarrow \left| { - x} \right| = 10$
$\Rightarrow x = - 10$
Now substitute the value of $x$ in the equation of $y$, which is $y = 2\left| x \right|$, as given below:
$\Rightarrow y = 2\left| x \right|$
As $x$ is negative, so $\left| {2x} \right|$ becomes $- 2x$, as given below:
$\Rightarrow y = - 2x$
$\Rightarrow y = - 2\left( { - 10} \right)$
$\therefore y = 20$
Hence the values of $x$ and $y$ are $\left( {x,y} \right) = \left( { - 10,20} \right)$
The values of $x$ and $y$ for $x > 0$, is $\left( {x,y} \right) = \left( {\dfrac{{10}}{3},\dfrac{{20}}{3}} \right)$
The values of $x$ and $y$ for $x < 0$, is $\left( {x,y} \right) = \left( { - 10,20} \right)$
Here while solving this problem for the values of $x$ and $y$, we must and should consider both the cases of $x$, when $x > 0$ and also $x < 0$, as the preferred case is not mentioned in the question. There are more important logarithmic basic formulas such as:
$\Rightarrow {\log _{10}}\left( {ab} \right) = {\log _{10}}a + {\log _{10}}b$
$\Rightarrow {\log _{10}}\left( {\dfrac{a}{b}} \right) = {\log _{10}}a - {\log _{10}}b$
$\Rightarrow$If ${\log _e}a = b$, then $a = {e^b}$
Hence $a = {e^{{{\log }_e}a}}$, since $b = {\log _e}a$