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

Hint: 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.\[\log {a^n} = n\log a\]