Question

If ${\log _2}\left( {x + y} \right) = {\log _3}\left( {x - y} \right) = \dfrac{{\log 25}}{{\log 0.2}}$ then find the values of x and y.

Answer 1

We will use the basics of logarithm here. First we will simplify the third term and then we will use it to find the value of x and y with the first two ratios.

Complete step-by-step answer:
Given that,
${\log _2}\left( {x + y} \right) = {\log _3}\left( {x - y} \right) = \dfrac{{\log 25}}{{\log 0.2}}$
Now
$\dfrac{{\log 25}}{{\log 0.2}}$
Here 25 can be written as square of 5 and 0.2 can be written in fraction form

$$\Rightarrow \dfrac{{\log {5^2}}}{{\log \dfrac{2}{{10}}}} \\\ \Rightarrow \dfrac{{2\log 5}}{{\log \dfrac{1}{5}}} \\\ \Rightarrow \dfrac{{2\log 5}}{{\log {5^{ - 1}}}} \\\ \Rightarrow \dfrac{{2\log 5}}{{ - 1\log 5}} \\\$$

Cancelling log5,
$\Rightarrow - 2$
Thus we simplified the last term. Now we will use it with the first two terms.
${\log _2}\left( {x + y} \right) = \dfrac{{\log 25}}{{\log 0.2}}$

$${\log _2}\left( {x + y} \right) = - 2 \\\ \Rightarrow \dfrac{{\log (x + y)}}{{\log 2}} = - 2 \\\ \Rightarrow \log (x + y) = - 2\log 2 \\\ \Rightarrow \log (x + y) = \log {2^{ - 2}} \\\ \Rightarrow \log (x + y) = \log \dfrac{1}{4} \\\$$

Cancelling log from both sides,
$\Rightarrow x + y = \dfrac{1}{4}$ →equation 1
Similarly using it with second ratio,
${\log _3}\left( {x - y} \right) = \dfrac{{\log 25}}{{\log 0.2}}$

$${\log _3}\left( {x - y} \right) = - 2 \\\ \Rightarrow \dfrac{{\log (x - y)}}{{\log 3}} = - 2 \\\ \Rightarrow \log (x - y) = - 2\log 3 \\\ \Rightarrow \log (x - y) = \log {3^{ - 2}} \\\ \Rightarrow \log (x - y) = \log \dfrac{1}{9} \\\$$

Cancelling log from both sides,
$\Rightarrow x - y = \dfrac{1}{9}$ →equation2
Now we will find the value of x in y form from equation2 $\Rightarrow x = y + \dfrac{1}{9}$
Putting this value in equation1 we get

$$\Rightarrow y + \dfrac{1}{9} + y = \dfrac{1}{4} \\\ \Rightarrow 2y = \dfrac{1}{4} - \dfrac{1}{9} \\\$$

Taking LCM,

$$\Rightarrow 2y = \dfrac{{9 - 4}}{{36}} \\\ \Rightarrow 2y = \dfrac{5}{{36}} \\\ \Rightarrow y = \dfrac{5}{{36 \times 2}} \\\ \Rightarrow y = \dfrac{5}{{72}} \\\$$

This is the value of y.
Now let’s find the value of x. Putting this value of y in equation2

$$\Rightarrow x = \dfrac{5}{{72}} + \dfrac{1}{9} \\\ \Rightarrow x = \dfrac{5}{{72}} + \dfrac{8}{{72}} \\\ \Rightarrow x = \dfrac{{13}}{{72}} \\\$$

Hence found the values of both $x = \dfrac{{13}}{{72}}$ and $y = \dfrac{5}{{72}}$.

Note: In this problem students just need to use simple logarithmic rules and apply them. Rules like,
1.${\log _b}a = \dfrac{{\log a}}{{\log b}}$
2.$\log {a^n} = n\log a$